Karl's Calculus Forum

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Current Forum Discussion Begins

i have a math prob. about airplane position, velocity.. 2 airplanes are in a straightline landing pattern and must keep at least a 3 mi separation. airplane a is 10 mi from touchdown, and is gradually decreasing its speed from 150 mph to 100 mph. Airplane B is 17 mi from landing and is decreasing its speed from 250 mph down to 115 mph. A) assuming the deceleration of each plane is constant, find the position functions s1 and s2 let t=0 represent the times when the aiplanes are 10 and 17 mi from the airport. B) find a formula for the magnitude of the distance d between the 2 planes. is it less than 3 at some time? if so find that time.
Rachel Vander Ploeg


lawton, mi USA - Saturday, April 12, 2008 at 16:13:21 (EDT)
Can you help solve my problem set. I already did number 3, but I can't solve the other numbers anymore. They're too difficult. Thanks a lot for your help. Download
rey pumaloy


USA - Thursday, March 20, 2008 at 00:42:11 (EDT)
ANy chance of an answer? Integration from +Infinity to -Infinity of the following e to the power of -x/SqRoot(4*pi*D*t) * sin (n*Pi*x/2) dx Set by my son so I would like to surprise Thanks
Richard


USA - Friday, March 14, 2008 at 11:28:37 (EDT)
The following Problem: A mass accelerated in a straight line from Zero V0 to Vend. What is the delta ΔkW consumption(indication on the kW meter)for every delta displacement(gain in distance ΔS of the mass???? Let Vm =intermetiate attained (staircaise)Velocity at attaind displacement, Let m=Mass, let Δv = delta velocity(gain in velocity per gain to the distance), let Δs = delta Displacemet(delta distance or incemental distance), let Δe = delta ΔkW power flow,

                Δv2
        (Vm2 + ---------- ) * m * Vm
                   2
Δe(kw) =------------------------------------
                Δs
IS THIS FORMULAR CORRECT?????
Gottfried Gutsche


Mississauga, Ont Canada - Wednesday, March 12, 2008 at 12:53:38 (EDT)
1. The problem statement, all variables and given/known data
1) If f(x)= sin^4x, then f '(pi/3)
2) Given f(x) = x/tanx, find f '(3pi/4)
3) If f(x) = sinxcosx, then f '(pi/6)
4) Differentiate: f(x) = x^2 + 2tanx
Question that I have answered but not sure if it's really the right answer:
5) Find the equation of the tangent line to the graph of f(x) = (x-1) / (x+1) when x = 1
2. Relevant equations
Product Rule?
* y= f x g
F'g + g'f

Chain Rule?
* y=f / g
* ((f'g) - (g'f)) / ((g^2))

Slope?
* y=mx+b

3. The attempt at a solution
1) f(x) = sin^4x then f '(pi/3)
Derivative of sinx = cosx therefore...
I'll assume that sin^4x has the derivative of cos^4x
Now, plug in the number...
cos^4(pi/3) = 0.065
Is that right?
2)
Given f(x) = x/tanx, find f '(3pi/4) Chain Rule:
((f'g) - (g'f)) / (g^2)
Therefore... ((1 * tanX) - (??? * X)) / ((tanx^2))
Let * be multiplication sign and the ??? to be the "I don't know".
So, I got stuck of what the derivative of "tanx". However, what I do know is that:
tanx = sinx / cosx
tanx' = cox / -sinx <-------Is that right?
If yes, then how am I suppose to make my equation by using "cox / -sinx "? By plugging that in... I get this:
(((1 * tanX) - ((cox / -sinx) * X))) / ((tanx^2))
Then I'm really stuck on that one... I mean, if I do plug in the "pi/3" to the "x" variables then it will just be a mess. Unless that's the only way to get the answer? Or should have I used the product rule instead?


3)
If f(x) = sinxcosx, then f '(pi/6)
Product rule:
y= f x g
F'g + g'f
Therefore...
= (cosx*cosx) + (-sinx*sinx)
= (cos^2x) + (-sin^2x)
= (cos^2(pi/6)) + (-sin^2(pi/6))
= 0.633
-Let * be a multiplication sign
-Is that right?

4)
Differentiate: f(x) = x^2 + 2tanx
So, I'll just get the derivative of the equation...
2x + 2(???)
-Let ??? be "I don't know".
So, I'm stuck. I have no idea what's the derivative of tanx have. I already encountered this problem in question #3 and I assumed that it would be:
tanx = sinx / cosx
tanx' = cox / -sinx
Is that right? If yes, then I would get this equation:
2x + 2(cosx/-sinx)
Is that right? If yes, can I simplify it much more?

5)
Find the equation of the tangent line to the graph of f(x) = (x-1) / (x+1) when x = 1
y=mx+b
Chain Rule:
= 1(x+1) - 1(x-1) / (x+1)^2
= x+1 -x +1 / (x+1)^2
= 2 / (x+1)^2
= Plug in "x"
= 2 / (1+1)^2
= 2 / 4
= 1/2
slope (m) = 1/2
Now that I have the slope, I'll just get the x & y values from plugging in "1" to the equation.
y = (1-1) / (1+1)
y = 0/2
y = 0
So: x = 1 and y = 0
Then reflect on the slope equation:
y = mx+b
Plug in the numbers from what I have gotten before:
0 = 1/2(1)+b
0 - 1/2 = b
-1/2 = b

So...
y = 1/2(x) + (-1/2)
I'll multiply the whole equation by "2" to make it more neater.
2y = 2(1/2x) + 2(-1/2)
2y = x -1

Is that right?


------------------------------------------
I know my solutions were kind of long. But I hope that you could help me. I really want to make this happen. Or at least answer the questions correctly or in a much simplified way. =)

PLEASE.
paola


Canada - Wednesday, March 12, 2008 at 00:48:16 (EDT)
Okay. I've been working on this problem for a solid two, front and back, notebook pages. Can someone please tell me if I'm on the right track with this? Differentiate: y=x^x 1. I took the natural log of both sides. ~ lny= xlnx 2. I used the formula, [lnu]= u'/u, for the xlnx ~ lny = x(1/x) 3. I " e'd" both sides to get rid of the natural log of y. ~ e^lny ******* 4. This is where I'm kind of stuck. I dont know which of the following should be the outcome of my "e-ing"... a. y= xe^(1/x) or b. y= e^(x(1/x) This could be an easy answer, but I'm a little confused. Could someone please help me?
Ariel


Cambridge, MN USA - Monday, March 10, 2008 at 21:41:29 (EDT)
Okay. I've been working on this problem for a solid two, front and back, notebook pages. Can someone please tell me if I'm on the right track with this? Differentiate: y=x^x 1. I took the natural log of both sides. ~ lny= xlnx 2. I used the formula, [lnu]= u'/u, for the xlnx ~ lny = x(1/x) 3. I " e'd" both sides to get rid of the natural log of y. ~ e^lny ******* 4. This is where I'm kind of stuck. I dont know which of the following should be the outcome of my "e-ing"... a. y= xe^(1/x) or b. y= e^(x(1/x) This could be an easy answer, but I'm a little confused. Could someone please help me?
Ariel


Cambridge, MN USA - Monday, March 10, 2008 at 21:39:45 (EDT)
Hi, A question. have you heard of the smoothstep function. So if you have heard of this function then this question is for you. Their exists a 3 degree polynomial function whose derivative is a solution to a smoothstep function. i.e.

f(x) = ax3 + bx2 + cx + d

when drawn the derivative of this function gives a curve as shown in image Fig 1. What is needed is a shown in Fig 2. Is it possible. How. ? PS, Since it is not possible to load an image here. Imagine Fig 1 as a less curvy 'S' and Fig 2 as a normal 'S' BRgds, kNish
knish


mumbai, MAH India - Friday, January 04, 2008 at 04:25:17 (EST)
Water is flowingat the rate of 6m^3/min from the tank shaped like a hemispherical bowl of radius 13m. Answer the following questions given that the volume of water in a hemispherical bowl of radius R is: V = (pi/3)y^2(3R-y) when the water is y meters deep.

a. at what rate is the water level changing when the water is 8in deep?

b. what is the radius r of the water's surface when the water is y in. deep?

c. at what rate us the radius r changing when the water is 8m deep?
Adam


kingston, pa USA - Sunday, November 11, 2007 at 00:38:43 (EST)
General info. I discovered a clear explanation of the Related Rate Problemin Calculus, J.Stewart., 4th Ed.
Robert


Dingmans ferry, PA18328General info: USA - Wednesday, November 07, 2007 at 15:19:40 (EST)
yx-1 = y1/2 * ln(yx-1) Is there a way to rearrange this equation is terms of x? i.e. x=?????? so I can solve it (without calculus) by just plugging in a y? Assume y >= 1, limit to natural numbers if necessary. I only need positive, real solutions for x. How do I solve for x when x is a root of a natural log? I think I could do it if there was only one x, or if there was no log. But with two x's and one in a log, I don't know how to rearrange this so x is by itself on one side of the equation.
David


Surrey, BC Canada - Thursday, October 25, 2007 at 19:31:00 (EDT)
I need some help on the following to determine whether or not they are continuous and explain why or why not: Is f(x) = 5x-10 continuous at x=2? Help would be greatly appreciated. Malga
Malga


USA - Sunday, October 21, 2007 at 15:55:49 (EDT)
I'm having problems trying to graph these conditions. I don't know where to begin. Please HELP!! 1. Sketch the graph of a function f(x) that satisfies the following conditions: a)f(-9)=f(1)=f(6)=0, b)f'(-4)=f'(3)=0, c)f"(-8)=f"(-1)=0, d)f'(x) positive only on (-4,3), e)f"(x) positive on (-8,-1) and (8,infinity). On graph, label all the points of inflection, local maximum, and local minimums.
NETTE


Denver, CO USA - Tuesday, October 09, 2007 at 03:43:30 (EDT)
given P = Y^tY - 2Y^tAC +C^tA^tAC Where Y is a column vector = a matrix with rows n and column =1. A is a matrix rows = n cols = 2 the first col are x, that is x1, x2,...,xn; the second col has 1 in each entry. C is a colum vector = a matrix with 2 rows and one column call the entries c1 & c2. ^t symbolizes transpose The equation for P reads in English: (Y transpose by Y) minus (twice times Y transpose by A by C) plus (c transpose by A transpose by A by C) I want to find the partial derivative of P with respect to c using the rules of linear algebra. I know the solution is -2A^tY +2A^tAC, in English minus (times A transpose by Y) plus (twice times A transpose by A by C) I am unable to locate a reference giving details on how to evaluate a derivative using the rules of linear Algebra with simple examples for me t work with. I have expanded written out P using n=2 and then found the partial derivatives using the normal rules for functions of two variables. It was most tedious and I would like to learn the linear algebra rules based on simple examples. Many thanks, Peter
Peter


Ireland - Monday, October 08, 2007 at 04:41:44 (EDT)
I am taking a distance learning Calculus course. I have been out of school for 15 years and was never very good at the maths in the first place. I am pulling my hair out trying to figure out a problem that should be simple. How do I approach it? Here it is: A car rental agency has rates of $75/day (with partial days charged at the full day rate) and $330/week. You are charged the daily and/or weekly rate, whichever is cheapest for you. For example, if you rented a car that was $100 daily or %50 weekly, the charge for 4 days use would be $400; the charge for 6 days use would be $550; the charge for 8 days would be $650 ($550 + $100) Let A(t) represent the average cost to rent the car for t days, where 0 < t < or = 15 I am confused and at a dead end.
Jordan Hyde


Springville, UT USA - Tuesday, September 25, 2007 at 18:05:58 (EDT)
[Question] Hi, my teacher gave us this problem, and he couldn't figure out why method was incorrect and why I got the answer I did. Given we know the gradient slope = <-56,1.886> at the point (2,0) on a surface f(x,y), in what direction, expressed as a unit vector, is f increasing most rapidly? [Difficulty] I solved the problem like this: Max slope = magnitude of gradient slope (gradient slope) dot (unit vector) = 56.03 -56.03Ux + 1.866Uy = 56.03 sqt(Ux^2 + Uy^2) = 1 Solving the system Ux= -.9977 Uy=.0672 The only problem is, by definition, I should be able to get the unit vector by taking the gradient slope vector and dividing by the magnitude of the gradient vector. or, <-56/56.03, 1.886/56.03> = = <-.9994,.0336> The weird thing is, the correct Uy value is almost exactly half of mine...what's going on??? [Thoughts] I tried a similiar technique for finding where the ∇f = 0 <-56, 1.866> dot (unit vector) = 0 -56.03Ux + 1.886Uy = 0 sqt(Ux^2 + Uy^2) = 1 Solving the system this time I got the correct answer, why here and not there?
John Smit


USA - Wednesday, September 19, 2007 at 00:05:15 (EDT)
l would like to join your program, to send and receive calculus problems. lam very good in math and l hope to enjoy your program with other students
stephen nyamapfumba


houston, texas USA - Saturday, September 15, 2007 at 19:52:41 (EDT)
Hi,

I have not done calculus in a while. Find the range for:

g(z) = 1 / square of 4-z^2

I can easily find the range for the other equations but I got stuck on this one since it's 1 / over a square, etc.

Thank you.
Martyn Reno


Montreal, QX Canada - Thursday, September 13, 2007 at 22:17:51 (EDT)
hey, can anyone help with solving equations by extracting square roots
varawdn
USA - Saturday, September 08, 2007 at 19:32:08 (EDT)
Can ε = 1/δ ? I mean such function should not have a limit, because as delta gets smaller epsilon gets bigger so function is not converging. Am I right?Thank you in advance very much.
felix


USA - Friday, July 20, 2007 at 11:38:04 (EDT)
A piece of wire 5 inches long is to be cut into two pieces. One piece is x inches long and is to be bent into the shape of a square. The other piece is to be bent into the shape of a circle. Find an expression for the total area made up by the square and the circle as a function of x. I need this to be as detailed as possible. I got the final formula...I don't know if it's right, or if I have to put it into the y=^#$*&^$^ form. So, if you could answer both of those questions, that would be EXTREMELY helpful.
Katie & Annamarie


USA - Thursday, July 12, 2007 at 15:03:35 (EDT)
Reply to Chase. If it is a simple graph here is what you need to think about: The graph of cos & sin repeat after 2pi or 360 degrees, the period The graph of cos & sin has a y distance from max to min of 2 units The graph of cos & sin is centred about y=0 look at your graph, think of it as a sin graph, what is its period? if it is 120 degrees then it is the graph of 3x if it is 720 degrees then it is the graph of 0.5x, call this number a. Does it pass through 0 and it it increasing to the right of 0? if not it is the graph of a sin but shifted to the right or left so it needs to be written as sin(x + - some number), call this number b, if a from above is not 1 you need to write this as sin(ax +- ab). What is the y distance from the max to the min? if it is not 2 then you need to multiply sin(ax +- ab) by some number C that is greater than or less than 1: Csin(ax +- ab). Is the graph centered about y=0? if not you need to add or subtract some number D to get Csin(ax +- ab) +- D. Hope this helps Peter
Peter


Ireland - Monday, June 11, 2007 at 09:06:21 (EDT)
How do you write an sine and cosine equation from a graph?
Chase


fort smith, ar USA - Tuesday, May 29, 2007 at 23:17:14 (EDT)
I need help with finding simultaneous tangent lines for two equations f(x)= ln x and g(x)= ex I have looked at the problem that was sampled but I am having troubles doing the substitution of f(x) and g(x) when trying to solve for one of the variables.
Josh


md USA - Tuesday, May 15, 2007 at 21:38:24 (EDT)
Reply to Mark: Just because an expression seems simple, it doesn't mean that it's easy to integrate. The antiderivative of your function cannot be expressed in any finite combination of elementary functions. Indeed you need to use Jacobian elliptic integral functions in order to integrate this. That is a topic that is usually not covered in first and second year courses.
Karl <Click to Send Email to Karl>
USA - Friday, March 30, 2007 at 15:25:35 (EDT)
This is a great site, and has kept me entertained for countless hours rediscovering my "A" level maths (I guess that equates to High School for those of you [no doubt the majority] on the other side of the pond!). I've set myself increasingly complex problems until I've finally stumped myself! I'm trying to integrate the following pretty simple expression:

f(x)=√(1+kcos(x))

I've tried the route of substituting x=arccos(w), x=arccos(w/k) and various mungeing ideas. I've tried expressing the resultant as logs, but no matter what twists and turns I try, I can't get it to fall out. If anyone has any clever ideas I would be very grateful to hear from you.

Many thanks,
Mark
Mark



Croyde, Devon UK - Thursday, March 29, 2007 at 10:23:13 (EDT)
I need help with calculus (integrate), out of school too long.
Charles Carter


Vine Grove, KY USA - Monday, March 26, 2007 at 14:02:26 (EDT)
Reply to Jesse: There is an almost identical problem to this in the online webpages. Please click here to see the worked solution. Then see if you can apply the same method to your problem.
Karl <Click to Send Email to Karl>
USA - Friday, January 05, 2007 at 12:53:05 (EST)
A spotlight on the gound shines on a wall 12 m away. If a man 2 m tall walks from the spotlight towards the building at a speed of 1.6 m/s,how fast is the shadow on the biulding decreasing when he is 4 m from the building?
Jesse


Los Angeles, California USA - Thursday, January 04, 2007 at 19:11:23 (EST)
02/11/2006 Math Squared learning program lets children from PreK-6th explore the whole new world of math, including computation, logic, critical thinking, problem solving, puzzle, real-life application, geometry, and other areas in math. Our founder believes that math is more than just the computation; children should be able to think out of the box and equipped to overcome further challenge. Our special Math Learning System teaches children more than just the math, it often helps other subjects. Math Squared, one of the best Math learning Systems, is now running a promotion for the licensing program. License fee is only $599, training and startup kit included. Multiple income sources. We supply workbooks and materials. Home classroom welcome. No experience required. Please visit www.m2-license.com for more info, and www.math-squared.com for Math Squared sample problems. kranti
kranti


USA - Saturday, December 02, 2006 at 01:36:25 (EST)
1/2f(-1,1)-2g(-3,3) when f(x,y)=x+2y-3, g(x,y)=4xy-54/x^2+3y
chrissy


tallahassee, fl USA - Friday, December 01, 2006 at 23:08:14 (EST)
I am to a find the equation to the tangent line of two parabolas. They all have the same slope. When I equate their derivatives my terms disappear. What do I do then? Need to find points of tangency of both parabolas F(x) =1-x², g(x) = -x²+8x-12
Monica


Sunrise, FL USA - Friday, November 24, 2006 at 14:46:05 (EST)
I have a math assignment due in two day on related rates and optimization problems and i am having soo much difficulty with them! anyone who could help me out i'd love you forever!! i'm stuck on my first question which is here: The base of a rectangular tank is 3m by 4m and is 10m high. Water is added at a rate of 8 m^3/min. Find the rate of change of water level when the water is 5 meters deep. Please, it's probably easy to someone out there.... thanks, ratch
Rachel


Canada - Tuesday, November 21, 2006 at 18:25:50 (EST)
In class we are doing a section about optimization problems. I understand that it's hard to set any specific rules about optimization because of the large number of potential problems, but are there any set equations that can be used in certain circumstances, i.e. finding the largest rectangle inscribed in a triangle, or cylinder inscribed in a cone. Any help with this subject would be greatly appreciated. My teacher doesn't teach, and expects us to learn it on our own :[
Nick


Philadelphia, PA USA - Wednesday, November 15, 2006 at 09:11:27 (EST)
Please could someone tell me the equations (either as a differential equation, or solved) for waves on water in Earth gravity? I seem to have found this (please correct me if I am wrong) 1) A wave train of one wavelength traveling in one direction is in the form of a curtate trochoid (see http://en.wikipedia.org/wiki/Trochoid ) with the sharper points up. Not a sinusoid. 2) As the wave height increases, the wave tends towards a cycloid, then tries to become a prolate trochoid, but this would force water to pass through other water, so the top of the wave breaks down into a line of foam. 2) Speed of waves varies as sqrt(wavelength). But eg. what happens when two or more wave trains interfere? Either when the trains are travelling in the wame direction, or in opposite diections, or at an angle to each other. Thanking you in advance.
Anthony Appleyard


England - Monday, November 06, 2006 at 02:42:07 (EST)
The following question is in the section of our Calculus text regarding the second fundamental theorem of calculus. I am having a difficult time resolving this question... If the integral of the function v(t-k)dt = S(t)+ C, where k and C are constant, what is S'(t)? Thanks.
Steve


East Stroudsburg, PA USA - Sunday, November 05, 2006 at 00:55:53 (EST)
Here is a simple problem that I don't seem to be able to come to terms with the wording of: "The cost for a truck driver is $7.50 per hour and the cost to operate the truck is 0.002v^2 where v is the average speed of the truck. How fast should the driver drive to minimize the total costs?" It seems obvious that one should set up some kind of cost function, differentiate it, set the derivative equal to zero to find the extremum (possibly check the sign of the second derivative to see if it matches what is wanted). However, how is "the cost for a truck driver is $7.50 per hour" to be understood? Are those wages he is paid, or some other expenses? In all, any suggestion how this is to be thought of? I just don't quite get what is meant by it.
Martin
USA - Friday, October 06, 2006 at 15:02:59 (EDT)
I've come across a problem in one of my Calculus books that I use in my AP Calculus class. It's in the related rates section, quite an early chapter in the book. The problem is basically as follows:

"Rain is falling into an inverted circular cone of height 10in. and top radius of 8in. The rain is falling at the rate of .1in3/min. However, the water in the cone is leaking out at the rate of 0.001h2. Will the cone ever overflow?"

It's easy to determine the basics that you need to work the problem, such as a formula for volume of a circular cone. I can easily see (and explain) that V' (or dV/dt if you would rather) is (.1 - 0.001h2). What is more difficult for me to see is how this can be solved using only limited, early-level Calculus (namely, related rates). To my eyes, this will likely end up requiring separable differential equations to solve.

Now, I've tried rewriting the volume in terms of the height only, using proportional triangles to relate the height to the radius of the cone. This I've then differentiated with respect to time and equated to the V' that is given in the problem, but that sets up the differential equation I referenced before.

I can tell from looking at the V' formula given that if the height of the water were to be 10in (the height of the cone itself), implying the cone is full of water, that the overall rate of change of volume would be 0. Intuitively I could see how a student would be tempted to state this means that the cone would actually fill to capacity and then never overflow. However, mathematically I'm not as convinced. Obviously V' refers to the instanteous rate of change of the volume with respect to time, so at that exact instant, there is no rate of change of water volume. Should that be satisfactory enough for me to conclude it will not overflow?

This problem "feels" like I'm making it far harder than it needs to be. I've looked at it from numerous angles, but I'm making no headway at this point. If someone could provide an idea or two, I'd be most appreciative.

Thanks,
Jason Karol
AP Calculus / Precalculus Instructor
Kathleen Senior High School
Lakeland, FL
Jason Karol


Lakeland, FL USA - Thursday, October 05, 2006 at 13:40:03 (EDT)
I am working on a calculus math project. I need to find the function and be able to explain using the below info. Can you help get started? Here's the problem. A company can only commit $4225 to this project without going into debt. Also, since the greenhouse can only hold so many trees, we need to make sure that we're spending no more that $100 per tree&emdash;otherwise we'll operate at a loss. Ed assured me this was possible&emdash;he even showed me how to do it&emdash;but he took most of his notes with him, and I'm no mathematical genius. Here's how the expenses add up. First, to build a greenhouse, you need $2,222 just in start up costs. Then, for each tree in the green house, you need $5 for a proper planter. But in addition, because the more trees there are, the easier it is to spread infection, the cost of disinfecting any one tree is $1 for each tree in the greenhouse. The one example Edgar did that I managed to save goes like this: suppose we plant 10 trees. Then we'd spend $2222 for the greenhouse, plus $50 for planters, plus $10 to disinfect each of 10 trees (meaning $100 for disinfecting the whole greenhouse). The total cost would be $2,372. So far, so good. But unfortunately this comes to over $237 per tree&emdash;that's bad. I tried figuring out what happens if we increase the number of trees, say to 100. In that case, I got the cost per tree to be a better (but still not acceptable) $127, but the total cost to be an exorbitant $12,722. I'm not even sure these figures are right, however. I'm sure you could tell me. I showed these figures to Sabbling, and she threw a fit&emdash;swore I'm going to wreck the company. She gave me three weeks to come up with a way to make this work, or I will be out of a job. I just know Jack is licking his lips over all this, waiting to take my place. Can you help me figure out what to do? I'm sure that Ed was sincere, just as sure I am that Jack is a duplicitous, two-faced, no good scoundrel. I have until September 22 to get on track with Sabbling (that's when the contract I signed says we have to place our order with the company). I need to take this info apply it to a calculus problem, using functions.
Sherri


Akron, OH USA - Wednesday, October 04, 2006 at 09:48:55 (EDT)
Reply to Allen: Graph the derivative on a piece of graph paper. Count the number of squares between the graph and the x axis from the origin to each x value you care to graph. But make sure you count squares below the x axis as -1 each and those above the x axis as +1 each. Graphing those counts will approximate what you want.
Karl <Click to Send Email to Karl>
USA - Friday, September 29, 2006 at 14:14:41 (EDT)
I was wondering if someone could give me advice on how to determine an approximate graph of a function given only the graph of the derivative? Thanks Allen
Allen


Holland, MI USA - Thursday, September 28, 2006 at 13:55:02 (EDT)
Reply to David: Finding the derivative of f(x) = xln(x):

Use the identity:

   bx = ex ln(b)
So what you really need to take the derivative of is
   eln(x) ln(x)  =  e(ln(x))2
which is nothing but an exercise in applying the chain rule (twice).
Karl <Click to Send Email to Karl>
USA - Thursday, September 21, 2006 at 18:40:18 (EDT)
Reply to David: You don't need to be able to integrate that nasty thing to get the answer to this. Let f(x) be the function under the integral. Let F(x) be an antiderivative of f(x). So F'(x) = f(x). If you take the integral from 1 to x3 of f(t) dt, then you get
   F(x3) - F(1)
and you want to take the derivative of this. Well the derivative of F(1) is zero because this term is a constant. For the derivative of F(x3), we apply the chain rule. You know that F' = f, so applying the chain rule gives f(x3)(3x2). Remember that f is the function under the integral. So I'll let you take it from here.
Karl <Click to Send Email to Karl>
USA - Thursday, September 21, 2006 at 18:35:23 (EDT)
I have two questions first how do you take the derivative of an integral?
f(x)=I(√(1+t3))dt from 1 to x3 
and how do you take the derivative of
f(x)= xln(x)
thanks David
David


USA - Tuesday, September 19, 2006 at 03:15:30 (EDT)
I need help integrating the following: x/(x^3+c^3). I know we have to use heaviside's method to split it into partial fractions but somewhow, I am not able to proceed. I could use any help granted to me. Thanks a lot in advance!!
Jalpan Dave


Jakarta, INDONESIA - Sunday, September 10, 2006 at 03:48:19 (EDT)
Looking with a rested eye at "Observe that x/(1+x) = 1 - 1/(x+1)" gives:

I =    x
     -----
     1 + x

     x + 1 - 1  (*add and subtract 1*)
  =  --------
     1 + x

     x + 1     - 1
  =  -----    -----
     1 + x    1 + x


     1     - 1
  =        -----
           1 + x

Peter
Peter


ireland - Thursday, August 24, 2006 at 05:08:32 (EDT)
Karl, You stated in your reply ot Jim:

 "Observe that x/(1+x) = 1 - 1/(x+1)"
How did you get that? It has got me stumped and I know that it should be simple. Thanks, Peter
Peter


Ireland - Wednesday, August 23, 2006 at 17:33:31 (EDT)
Reply to Jim: Observe that x/(1+x) = 1 - 1/(x+1). Now integrate each term separately.
Karl <Click to Send Email to Karl>
USA - Monday, August 14, 2006 at 12:58:50 (EDT)
How do I integrate x / (x+1)?
Jim
USA - Saturday, August 12, 2006 at 11:49:50 (EDT)
Reply to Ashlie: Remember that 16 = 24. So 163x = (24)3x. Remember the rule for taking a power of a power -- you multiply the exponents. So (24)3x = 212x.
Karl <Click to Send Email to Karl>
USA - Monday, August 07, 2006 at 12:56:07 (EDT)
Hi everyone, Thanks for readin g this in advance. I am starting to go back to school after being in naval nuclear engineering and was given credit for calculus 1 and must take calc 2 and 3 to finish my degree. I unfrotunately forgot most of everything. I have been looking at software to help freshen up on and am wondering if anyone has used Pro one ADvanced Mathematics or Calculus Wiz to help learn. I read a lot on this page and it starting to come back to me a bit but need some exercises so that I dont jump into calc 2 without no clue. Thanks for your help, Todd Barnett
Todd Barnett


Albany, GA USA - Tuesday, August 01, 2006 at 15:46:42 (EDT)
HOw do you solve this:
16^3x , base 2
Directions: re-write the exponential expression to have the indicated base.
Ashlie


USA - Tuesday, August 01, 2006 at 09:33:21 (EDT)
Reply to Signifier: The integral of dx/x is ln|x| + C. As x goes to zero, ln(x) goes to -∞. This means that you cannot take this integral with zero as either the upper or lower limit. You also cannot take this integral over any real interval that includes x = 0.

Perhaps the material you were looking at referred to doing this integral in the domain of complex numbers. There you can choose your path of integration arbitrarily between any two points you choose as limits of the integration. So, for example, you could integrate this from -1 to 1, which would include x = 0. But in the complex domain you can detour around the troublesome point (which is called a pole of the function). Your detour can come arbitrarily close to the pole, but it must bypass it either to one side or the other. Depending upon which side you bypass it, this will add a term of ±πi, where i is the square root of -1.
Karl <Click to Send Email to Karl>
USA - Friday, July 21, 2006 at 13:04:02 (EDT)


I recently encountered something in a problem, and I have no idea what to do.

What I am wondering is, are the following two things equivalent:

Integral from 0 to 5 of dx / x

and

Integral from 0 to 5 of dx / (x + dx)

?

It seems intuitively plausible to me that they are: After all, dx is going to 0 in the limit. But is this true? Is there any strong mathematical foundation behind just chucking the "+ dx" right out of there... or do these two expressions represent two very different things...? I thought about making a u substitution: let u = x + dx, then (du/dx) = (dx/dx) + d(dx)/dx... but what is the "derivative of a differential" equal to ? 0?

Can anyone help me here? I am excited and confused.
Signifier
OR USA - Monday, July 17, 2006 at 22:00:55 (EDT)
Find f'(x) for
F(x) = x2 + 2x
       ------------------
       √3x+1

Without using the Chain Rule or the Difference Quotient

I hope you can help:)

Thanks
Angie


MN USA - Wednesday, May 24, 2006 at 16:26:13 (EDT)
Sudheendra, Are R, T, a & b constants or do they change with v? What does this equation refer to, as if R, T, a & b change with v there might be some arrangement that is a constant? Peter
Peter


Ireland - Wednesday, May 24, 2006 at 08:20:34 (EDT)
hi i had problem in doing some integration n differentiation for my project. pls help me in this. im typeing the integRal required. I = integration((1/(v^2+2*b*v-b^2))-(R*T*v/((v-b)^2))+(a*v*(2*v+2*b)/((v^2+2*b*v-b^2)^2))) with respect to dv. thanq
Sudheendra


Mangalore, Karnataka india - Wednesday, May 24, 2006 at 06:45:17 (EDT)
Angie, Does your textbook give a definition for a cusp? Have you tried to sketch the graph of the function? You should try a sketch and then try to draw a few tangents to see if there are any places where the tangent is vertical and also to see if there are any places where you could draw two or more lines that appear to be tangents at the same point. Use those x values and plug them into the derivative of f(x) and see what happens. Peter
Peter


Ireland - Tuesday, May 23, 2006 at 14:27:55 (EDT)
Hello, I'm having difficulty determining if a function has a vertical tangent or cusp. Can someone please explain how to find theses? The function I'm working with is f(x) = x^(3/4)÷(x+2). I appreciate any help, as my textbook doesn't explain how to find these. Thanks! Angie
Angie


Canada - Tuesday, May 23, 2006 at 12:13:09 (EDT)
Brad, What do you get for
d/dy(ex2)
Peter
Peter


Ireland - Tuesday, May 23, 2006 at 06:10:24 (EDT)
i need help passing my calc course maybe if u could give me some reference material it would help
wendell M


USA - Friday, May 12, 2006 at 17:59:00 (EDT)
How do you integrate e2x/ex-3
Fiffy


USA - Friday, May 12, 2006 at 09:27:29 (EDT)
Hi; I'm having trouble with this problem: Find the derivative of
y = x2ex2.
Mainly unsure about the rules concerning the derivation of ex in general... I know as far as this:
x2 × (d/dy(ex2) + ex2 × 2x
However, I'm rather uncertain as to where to go afterwards because I'm not arriving at the given answer, which is
2xex2(x2 + 1)
Thanks in advance, Brad
Brad H.


Sterling Heights, MI USA - Sunday, May 07, 2006 at 12:55:29 (EDT)
Find the volume of the solid generated by revolving the region bounded by:
x2/a2 + y2/b2 = 1

- about the x-axis
- about the y-axis

x>=0, y>=0 (I think, if I remember the problem correctly!)
Has to do with integration to solve it (obviously). Thanks - trying to help my kid with homework, altho I was a math major, it was too many years ago!
Joan


USA - Thursday, May 04, 2006 at 11:50:50 (EDT)
I need help with a word problem for my Calculus for Business & Economics class. In planning a small restaurant, it is estimated that a profit of $5 per seat will be made for the first 80 seats. On the other hand, the profit on each seat will decrease by 5 cents for each additional seat above 80 Find the number of seats that will produce the maximum profit and the miminum profit.
Trina Vidal


Frisco, tx USA - Wednesday, May 03, 2006 at 23:43:57 (EDT)
Reply to Antony: See this page for general treatment of acceleration, and scroll down to see your problem (constant acceleration) in particular.
Karl <Click to Send Email to Karl>
USA - Tuesday, April 25, 2006 at 14:55:27 (EDT)
I have a problem with a problem. Maybe someone can help me? Show that for motion in line with constant acceleration a,initial velocity, and initial displacement, the displacement after time t is
antony


murray, ky USA - Monday, April 24, 2006 at 23:15:03 (EDT)
Reply to Elizabeth: For details of the cone-in-the-sphere problem click here.
Karl <Click to Send Email to Karl>
USA - Sunday, April 16, 2006 at 14:34:28 (EDT)
Reply to Yvonne: So you have two pieces, one of length, x, and the other of length L-x. The radius of the circular piece is x/2π. The square has length-of-side, (L-x)/4. So the total area is

        (L-x)2
  A  =          +
          16
  x2
    
   4

So taking the derivative

  dA     2(L-x)
      =          +
  dx       16
  2x
      =  0
   4

Note that I have set this to zero in order to find the optimum. Solve for x.
Karl <Click to Send Email to Karl>
USA - Sunday, April 16, 2006 at 14:31:03 (EDT)


i'm using calculus to find the maximum volume of a cone inside a sphere of radius 5cm. ive got 2 write the (h) height and the (r) radius from the volume of a cone equation in terms of x. and hence express the (v) volume in terms of x. where do i start? xx
elizabeth


west y, UK - Thursday, April 13, 2006 at 17:02:36 (EDT)
This should be straight forward but I cna seem to get a handle on it. Can anyone out there help? A piece of string at lenght L cm is cut into 2 pieces. One piece of length X is made into a circle and the other into a square. A) Find the value of X that makes the sum of the areas a maximum. THIS IS INTUITATIVE - DON'T CUT THE STRING - LET THE ENTIRE LENGTH FORM A CIRCLE. b) Find the value of x that minimizes sum of the the areas. I CAN'T SEEM TO GET STARTED ON SETTING UP THE RELATIONSHIP BETWEEN THE EQUATIONS. CAN YOU HELP?
Yvonne


tucson, az USA - Tuesday, April 11, 2006 at 19:44:50 (EDT)
Reply to Andre: I'm assuming you have already eliminated h by substituting the Pythagorean expression that gives h in terms of r. So now you have the volume equation in terms of r and the constant, R.

If you have

  r  =  R -
 θR
   

then you can substitute the right-hand expression into the volume equation everywhere you see r. You should end up with a volume equation in which the only independent variable is θ. The symbol, R, will still be there, but it, remember, is a constant. Now find dV/dθ by taking the derivative of the volume equation with respect to θ. Whatever experssion you get for dV/dθ, set it equal to zero. Now you have an equation in which the only unknown is θ. Solve for θ and you're done.
Karl <Click to Send Email to Karl>
USA - Saturday, April 08, 2006 at 11:16:12 (EDT)


Hey Karl, When you stated "Use the relationships already shown to first substitute for h using r, then to substitute for r using θ" did you mean to use r, when
r=R-((θ R)/2π)
in the volume formula where "h"? I got all of what you said except for the substitute part. You kinda lost me there with your wording, but I really appreciate the help so far!
Andre
Houston, TX USA - Saturday, April 08, 2006 at 04:44:38 (EDT)
Reply to Andre: The distance from the apex to the hem of the cone will be R, where R is the radius of circle of filter paper that you start with. The radius, r, and the height, h of the cone will be related by:
   R2  =  r2 + h2
or equivalently
   h = √R² - r²
To see why this is the case, picture the cone in crossection. You will see that the relationship given above is simply the Pythagorean formula.

If the wedge you cut out of the paper is angle θ radians, then the length of circumference removed is θR. That means that what's left is 2πR - θR. This will be the circumference of the base of the cone (picture in your mind what happens when you form the cone from the paper). Dividing that circumference by gives the radius of the base of the cone, r.

  r  =  R -
 θR
   

The volume of a cone is given by

  V  =
 1
  
 3
 π r2h

Use the relationships already shown to first substitute for h using r, then to substitute for r using θ. Now take the derivative, dV/dθ, set it to zero, and solve for θ (remembering that the radius, R of the original filter paper is a constant).
Karl <Click to Send Email to Karl>
USA - Friday, April 07, 2006 at 13:25:11 (EDT)


Reply to Julius: 1) The problem states that
  dV
      =  0.1 ft3/min
  dt
The volume of the sphere is given by V = (4/3)πr3. Taking the derivative of that with respect to t gives

  dV
    
  dt
  =  4 π r2
 dr
   
 dt

You can find r by using the volume equation replacing V with 1/2 ft3 and solving for r. You know dV/dt from the first equation. Solve for dr/dt.

2) The volume of a cylinder is V = π r2h. The radius, is given as 20 feet. Again dV/dt is given. Taking the derivative of the volume equation, assuming fixed radius:

  dV
    
  dT
  =  pi r2
 dh
   
 dt

Solve for dh/dt.

3) The volume of the cylinder is, again, V = π r2h. Note that V is given, so you can use that to find h in terms of r. The cost function is (two circles of radius, r, at $10 per square meter plus a rectangle that is 2πr by h at $8 mer square meter).

  C  =  $10 2π r2  +  $8 2π rh
Substitute the expression for h in terms of r into the above. Then take the derivative, dC/dr, set it to zero, and solve for r. The back-substitute to find h.
Karl <Click to Send Email to Karl>
USA - Friday, April 07, 2006 at 13:04:23 (EDT)
Question on how to start this problem. I don't know if it is ask to make use of a circle or a cone. Someone wants to make s Melita style coffee filter from a round piece of coffee filter paper. They want to cut out angle x and glue the two edges of the wedge to do so. 1)Find a formula for the volume of the resulting cone as a function of the angle x. 2)How should they choose the angle x if they want the volume of the resulting cone to be as big as possible. Ok, I know they want to use a circle, but I'm confused after that part.
Andre


Houston, TX USA - Friday, April 07, 2006 at 10:34:32 (EDT)
Hey, i have a few word problems that I'm having trouble with. 1) A little boy buys a spherical balloon of total volume 1 cubic foot. he starts blowing to fill the ballon at a rate of .1 cubic feet per minute. how fast is the radius of the balloon increasing when he has the balloon halfway blown up? 2) A cylindrical swimming pool is being filled from a fire hose at a rate of 5 cubic feet per second. if the pool is 40 feet across, how fast is the water level increasing when the pool is half full? 3) A cylindrical can is to hold 20 (pie) m^3. the material for the top and bottom cost $10/m^2 and material for the side is $8/m^2. Find the radius and height of the most economical can.
Julius


houston, tx USA - Thursday, April 06, 2006 at 17:27:11 (EDT)
Reply to Wendy: Suppose p/q = log{base 2}(7). Taking 2 to the power on both sides:
  2p/q  =  7
where p and q are both positive integers. Raise both sides to the q power:
  2p  =  7q
Consider that 2p has a prime factorization of nothing but 2's and 7q has a prime factorization of nothing but 7's. How can those both be equal to the same integer?
Karl <Click to Send Email to Karl>
USA - Sunday, April 02, 2006 at 16:43:11 (EDT)
Prove (by contradiction) that log(base 2)7 is irrational. help please!!! Wendy
Wendy


AUS - Friday, March 31, 2006 at 23:32:21 (EST)
Reply to Blake: Picture it in cross section. That would be a rectangle inscribed in a circle. If the upper right corner of the rectangle is at (x,y) and the circle's radius is r, then
   y = √r² - x²
If the y-axis is also the axis of the cylinder, then x is the radius of the cylinder and 2y is its height. Presumably the "best fit" would be to maximize the volume of the cylinder. The volume of a cylinder is given by V = π R2h, where R is the radius of the cylinder and h is its height. Replacing R and h with the expressions given for them above:
   V  =  2π x2r² - x²
Now take the derivative, dV/dx, of the above. Set it to zero and solve for x. You have to apply the product rule.
   0  = 4π x √r² - x²  -  2π x3 / √r² - x²
Multiply through by r² - x² and divide by 2π to get
   0  = 2x(r2 - x2) - x3
Divide this by x and solve the resulting quadratic. Back-substitute to find y.
Karl <Click to Send Email to Karl>
USA - Tuesday, March 28, 2006 at 13:19:38 (EST)
I need to find out how to inscribe a cylinder that is inside a sphere. The sphere has square root of 3 as its radius and i need to find out how to find the base and height of the cylinder that best fits the sphere. Asked my teacher but he says "I do not have enough time for you" please help.
Blake


Ont Canada - Sunday, March 26, 2006 at 22:30:57 (EST)
I need to differentiate: f(t)=(1 + ln t )/ (1 - ln t )
Keith


FT Hall, Id USA - Monday, February 27, 2006 at 12:48:47 (EST)
Cancel the first problem on there. I had a "brain-flatuation". Because the cross section is a square (not a rectangle) I should have

2int(f(x)*f(x)) between 0 and 2, instead of (x*f(x))

Still need assistance with the second one however.
Tony



Phoenix, AZ USA - Wednesday, February 22, 2006 at 13:53:28 (EST)
I'm trying to get an explanation to a couple of the sample problems from the AP Calculus Practice Book at the college board website. They are both on page 30 in the following pdf link.

http://apcentral.collegeboard.com/repository/05836apcoursdesccalc0_4313.pdf

First is #22. The base of a solid S is a semicircular region enclosed by the graph of y=sqrt(4 - x2) and the x-axis. If the corss sections of S perpendiculr to the x-axis are squares, then the volume os S is?

I set this up as 2*int(x*f(x)dx) between 0 and 2. I keep getting half of what they get.

Next is Question 24.

                pi*ex
If f'(x) = sin( ---- ) and f(0)=1, then f(2) =
                  2

Having difficulty getting started and any pointers as to the right direction would be appreciated. Also, I could not find any solutions to these online. If anyone has any links, they would be appreciated as well.
Tony


Phoenix, AZ USA - Wednesday, February 22, 2006 at 13:37:34 (EST)
Keith, You have to use the quotient rule which is the ((bottom function multiplied by the derivative of the top fcn) minus (the top fcn multiplied by derivative of the bottom fcn)) (all divided by the bottom fun squared). Are you familiar with the prime notation? (1 + sin x)' means the first derivative of (1 + sin x) which is (cos x). The best way to tackle this is to call the top functin U for upper and the bottom fcn V then the quotentint rule is for y = U/V is y' = ((VU') - (UV'))/(V^2). Hopefully this will get you started, if you need further assistance please post your work and I will comment on it. Peter
Peter


Ireland - Wednesday, February 15, 2006 at 18:02:32 (EST)
My question pertains to text found on this page: http://www.karlscalculus.org/calc1.html#s1_4 At the end of the second paragraph in section 1.4, we find the text: "... on all subsequent days the amount will be that close ...". From my point of view, this phrase requires that there be two consecutive days on which the amount of milk delivered is the same. But this cannot be true if the amount delivered on day n is 1+1/n gallons. For example, let's say that on a particular day I want the amount delivered to be 1.01 gallons. So, on day 100 I get 1.01 gallons. Now, the phrase above says that: "... on [a] subsequent [day] the amount will be that close ...". I think that any subsequent day will be associated with an amount that is LESS than 1.01 gallons. Am I mistaken? ~ Mark
Mark Morse


Seattle, WA USA - Wednesday, February 15, 2006 at 17:37:15 (EST)
I am trying to differentiate this: Y = (1 + sin x)/(x + cos x) I was trying to use the quotient rule but am not sure if i performed the right steps.... and Y = (x)(sin x)(cos x) I didn't know where to start
Keith


ft hall, id USA - Wednesday, February 15, 2006 at 15:31:53 (EST)
Keith, Please put brackets around the equation to indicate the order. For example is it y = (1+sinx)/(x+cosx) or y = ((1+sinx)/x)+cosx or y = 1+(sinx/x) +cosx and so on. Peter
Peter
Ireland - Wednesday, February 15, 2006 at 13:07:09 (EST)
trying to differentiate y = 1+sinx/x+cosx
keith


fort hall, id USA - Tuesday, February 14, 2006 at 18:11:48 (EST)
trying to differentiate y = 1+sinx/x+cosx
keith


fort hall, id USA - Tuesday, February 14, 2006 at 17:44:57 (EST)
Karl, Have you got any information regarding the Wronskian and Abels formula in the solution of 2nd order differential equations and also the proof of the use of the Wronskian. I am at sea with this at the moment and hopefully I will be able to come back with more specific information. Regards, Peter Gibney
Peter Gibney


Ireland - Friday, February 10, 2006 at 15:11:20 (EST)
I have a question that if you can help that would be wonderful. It's a chemistry problem which deals with 2nd order reaction rates. I have to integrate:
             dx
       --------------
         [A-x][B-x]
which somehow I'm supposed to come up with:
    1         [B - x][A]
--------- ln -------------
[B] - [A]     [A - x][B]

I tried doing partial fractions, but I had in the ln just the
    [A - x]
ln -------------
    [B - x]

which doesn't correspond to what I'm supposed to get and there isn't even a [A] or [B] in the ln either! If you can help me solve this "mystery" that would be great :) Thank you so much!
Matthew


Davis, CA USA - Sunday, January 29, 2006 at 04:09:46 (EST)
Suppose that S(q) is the price per unit (in dollars) of widgets which will induce producers to supply q thousand widgets to the market, and suppose that D(q) is the price per unit at which consumers will buy q thousand widgets A. Which is larger , S(100) or S (150), and why(be wordy)? B. Which is larger, D(100) or D(150), and why ( be wordy)? C. If D(100)=10 and S(150)=10, what will you predict about the future selling price widgets (currently at 10$)? Justify your prediction
Malcolm Vella


Philadephia, PA USA - Thursday, January 26, 2006 at 18:36:29 (EST)
Two questions about function continuity: 1) if a function is continious everywhere, its limit at minus infinity and plus infinity are both 0, does it necessarily have a maximum? and if the function is non-negative everywhere, what then? 2) how do you prove that if a function is continuious between point A and infinity and its limit at infinity is a constant L then that function is uniform continuious between A and infinity? Appreciate your help, Ron.
Ron


Tel Aviv, Israel - Thursday, January 26, 2006 at 16:00:01 (EST)
I'm wondering if anyone can help me with the following question. I have already completed it and handed it in but I am now preparing for my exam and feel that being able to do a question like this would help. If someone could give me a step by step tutorial it would be greatly appreciated. A construction company has been offered a build-operate contract for $7.8 million to construct and operate a trucking for five years to transport ore from a mine site to a smelter. The smelter is located on a major highways and the mine is 3 km into the bush off the road. The gravel road will not ber perpendicular to the highway. Construction (capital) costs are estimated as follows: Upgrade to the highway (i.e. repaving) will be $ 200 000/ km New Gravel road from mine to highway will be $500 000/km Operating conditions are as follows: There will be 100 return trips each day for 300 days a year for each of the five years. Operating costs on the gravel road will be $65/h and the average speed will be 40 Km/h Operating costs on the highways will be $50/h and the average speed will be 70 km/h. Use calculus to determine if the company will accept the contract and the distances of the paves and gravel road sections producing optimum conditions (max. profit). What is the max profit? Do not consider the time value of money.
Mark


Minesing, Ont Canada - Wednesday, January 25, 2006 at 21:07:29 (EST)
Could someone tell me what a limit is? I just dont quite get it. Like what is the basic definition or concept of limit
Kobe
USA - Tuesday, January 24, 2006 at 12:04:18 (EST)
Hi, Regarding my request for assistance with:

             1  dx
integrate -----------
          (x2 + a2)3/2

I have now figured out where I was going wrong and I cave sucessfullly carried out the integration. Best regards, Peter Many thanks, Peter
Peter


Ireland - Sunday, January 22, 2006 at 17:40:09 (EST)
Hi, Regarding my request for assistance with:

             1  dx
integrate -----------
          (x2 + a2)3/2

I have now figured out where I was going wrong and I cave sucessfullly carried out the integration. Best regards, Peter Many thanks, Peter
Peter


Ireland - Friday, January 20, 2006 at 16:26:06 (EST)
Hi, I would be obliged if I could get some assistance with an integral that has me really stumped:

             1  dx
integrate -----------
          (x2 + a2)3/2

Many thanks, Peter
Peter


Ireland - Friday, January 20, 2006 at 14:37:01 (EST)
Reply to Michelle: Let r be the radius of the cylinder and let h be its height. Then your constraint equation is the volume equation:

  V  =  24π inches3  =  πr2h

From this you can easily get h in terms of r:

  h  =
   V
     
  πr2

Your cost function is proportional to (and we'll just say it's equal to) the area of the sides (2πrh) plus three times the area of the bottom (3πr2).

   C  =  π (2rh + 3r2)

Now replace h with its equivalent from the constraint equation:

   C  =  π
 
 2rV
     +
 πr2
 3r2
 
  =
  2V
      +
   r
  3πr2

Now you find dC/dr of this, set it to zero, and solve for r.

  dC
      =  0  =  -
  dr
 2V
     +
 r2
  6πr

Substituting 24π inches3 for V (as that's what volume is given in the problem), applying the commutative law, and multiplying through by r2:

  0  =  6πr3  -  48π inches3

which you can solve easily for r. Back-substitute the solution for r into the constraint equation to get h.
Karl <Click to Send Email to Karl>
USA - Tuesday, January 10, 2006 at 12:43:47 (EST)


Hi i need help with this question: A circular cylinder container, open at the top and of capacity 24(pie) cubic inches, is to be manufactured. If the cost of the material used for the bottom of the conatainer is 3 times that used for the curved part, and if there is no waste of material, find the dimensions that will minimize the cost. thanks in advance
Michelle


Toronto, Canada - Monday, January 09, 2006 at 00:33:28 (EST)
Sorry - here we go:

Couldn't get that mad email link to work so here is a temp mail of mine: bromfieldcourtpathway@hotmail.com

Cheers!
Mason
London, UK - Tuesday, January 03, 2006 at 19:13:03 (EST)


Hi, I am working on my maths to gear up for an Economics masters I am starting this Autumun. One of the questions I am working on has got me stumped though. The text says that:
             Q2
            ----
             10
should differentiate to equal:
             Q
            ----
             5
I cannot seem to arrive at this answer - can anyone assist????


Best Regards, Mason
Mason
London, UK - Tuesday, January 03, 2006 at 18:57:33 (EST)


Reply to Bill Awesome: If by "standard part" you mean the limit as H goes to infinity, then you got it right. I've never seen that nomenclature before.
Karl <Click to Send Email to Karl>
USA - Tuesday, December 20, 2005 at 12:25:00 (EST)
Find the standard part of:
(H4+3H2+1) / (4H4 + 2H2 +-1)

I multiplied it by H-4, and I then found the standard part to be 1/4. Am I correct?
Bill Awesome



USA - Monday, December 19, 2005 at 23:59:17 (EST)
Reply to flexx: The surface area is given by the integral

 
 r ds

In your problem, r = f(y) (because you are rotating about the y-axis), and ds = √1 + (f'(y))² dy. Since you have f(y) = sin(y), it means that (f'(y))² = cos²(y). So the integral is

 π
 
 0
 sin(y) √1 + cos²(y) dy

Let u = cos(y). Then du = -sin(y) dy, and the integral becomes

  -2π
 -1
 
 1
1 + u² du

which can be integrated by substituting tan(v) = u, or by hyperbolic subsitution.
Karl <Click to Send Email to Karl>
USA - Sunday, December 18, 2005 at 18:25:06 (EST)


please help me to solve this integrals of surface area , i tried many times but i cant x = sin y (0) les than or equal ( Y )less than or equal (pi) revolved around y -axis please help me
flexx


USA - Sunday, December 18, 2005 at 14:01:10 (EST)
Reply to Jana: If you have two points, (x1,y1) and (x2,y2), then the distance between the two point, s, is given by:
  s2  =  (x1 - x2)2 + (y1 - y2)2
or equivalently
         _______________________
  s  =  √(x1 - x2)2 + (y1 - y2)2

Karl <Click to Send Email to Karl>
USA - Friday, December 16, 2005 at 20:46:54 (EST)
What is the equation for the "distance formula"?
Jana


Orange, Ca USA - Friday, December 16, 2005 at 19:43:19 (EST)
Reply to Sara: In problems where you have to find the limit as x goes to infinity of the quotient of two polynomials of the same degree, the answer is always the ratio of the leading coefficients. I'll do your example of

   lim
  x → ∞
  3x - 1
        
   x - 2

by the rule I indicated you should get a limit of 3. You can see why if you substitute 1/u = x and take the limit as u goes to zero.

   lim
  u → 0
  3/u - 1
         
  1/u - 2

Now multiply top and bottom by u.

   lim
  u → 0
   3 - u
        
  1 - 2u

You can see that as u goes to zero, this limit goes to 3. In the other problem where you have the limit as x goes to infinity of a ratio of two fourth degree polynomials, try the same thing. Replace x with 1/u and take the limit as u goes to zero. In this case you will multiply top and bottom by u4 to simplify the limit.
Karl <Click to Send Email to Karl>
USA - Tuesday, December 13, 2005 at 18:59:07 (EST)


lim x as it approaches ∞ 4x^3+2x^2-3x+3/2x^3-3x^2-7x-6 work done so far 4x^3/2x^3=2
sara


USA - Monday, December 12, 2005 at 15:46:07 (EST)
lim x-∞ 3x-1/x+2 work done do far 3x/x-1/x/x/x+2/x
sara


USA - Monday, December 12, 2005 at 15:40:31 (EST)
Hey guys, I'm trying to make a web site for tutors. I want to figure out what things are difficult for them so I can help them. Please take my 5 question survey below! Go here: http://www.questionpro.com/akira/TakeSurvey?id=319354 Thanks! Ryan
Ryan
Dallas, TX USA - Monday, December 05, 2005 at 23:39:37 (EST)
Reply to Gabe: You are the first person to come to me with a partial diff-eq ever. I don't know how much of the methodology you understand, so I will start at the beginning and provide a mini-lecture on the subject.

Your heat equation is looking for a function, T(x,t), which makes it an equation in time and in one spacial dimension. So it could be a staight bar, or as you indicate, a slab where temperature is uniform over the areas of each of the two surfaces.

  ∂T
      =  A
  ∂t
2T
     
  ∂x2

First thing -- before applying boundary conditions -- is to get a general solution to this. We do this by imagining that the solution is the product of two functions, one purely a function of time, the other purely a function of space. So we let T(x,t) = F(x)G(t), and apply the partial differential equation to the product.

  F(x)G'(t)  =  A F"(x)G(t)
Now we can solve time and space equations separately. Solving the time equation first, observe that F(x) and F"(x) are constants with respect to time. If you let μ = -F"(x)/F(x), then the time equation is
  G'(t)  =  -Aμ G(t)
to which the solution is G(t) = C e-Aμt, where C is a constant. Observe that no matter what C is, the ratio, G'/G = -Aμ remains the same. Now we do the space equation. Note that you can replace G'(t) with -Aμ G(t) in the space equation to simplify it. When you do, the A G(t)'s divide out leaving
  -μF(x)  =  F"(x)
the solution to which is F(x) = Mcos(√μx) + Nsin(√μx), where M and N are arbitrary constants. Letting ω2 = μ, this becomes F(x) = Mcos(ωx) + Nsin(ωx). This gives T(x, t) = e-Aω2t (Mcos(ωx) + Nsin(ωx)), which is the general solution to the one dimensional heat equation.

The rest of the problem is to apply the initial condition and boundary conditions to establish the eigenvalues that work for ω and the weights, M and N for each eigenvalue. With a heat equation we do the latter by applying the boundary conditions to establish the steady-state solution, T(x, ∞). The steady state solution has the property of ∂T/∂t = 0, which implies that 2T/∂x2 = 0. This means that T(x, ∞) must be a linear function of x.

   T(x, ∞)  =  mx + b
We apply the boundary conditions to find m and b. The right-hand boundary condition is ∂T/∂x = KT, which requires that m = K(m + b). The left-hand boundary condition requires that b = 1. So

  m  =
   K
     
  1-K

Now we find the eigenvalues, eigenfunctions, and their weights. We do this by taking the difference between the initial condition and the steady state solution and extracting a Fourier series from it.

  T(x, 0)  -  T(x, ∞)  =
   K
      x
  K-1

The left-hand boundary condition requires that we use only sin(ωx) terms to ensure that the eigenfunctions are zero at the left boundary. The right-hand boundary condition requires that ω be odd multiples of π/2 to ensure that the derivative of the eigenfunctions are zero at the right boundary. So ωi = (2i+1)π/2 are the eigenvalues, with i being integers from zero to infinity. The eigenfunctions are

  Ei(x)  =  sin(((2i+1)π/2)x)
Multiply this by the difference between the initial condition and the steady state, and integrate the result from x=0 to x=1. This will give you the Fourier weights, Ni. Note that since you are integrating over only a quarter wave, you need to scale the Fourier integral by a factor of 4 to get the correct values for Ni.

The full solution is

  T(x, t)  =  T(x, ∞) +

 i=0
 Ni e-Aωi2t sin(ωix)


Karl <Click to Send Email to Karl>
USA - Friday, December 02, 2005 at 09:13:04 (EST)


Hi, I am trying to solve the heat equation for a slab. Slab thickness, x, varies between 0 and 1, and temperature T varies between 0 and 1. t is time. Any help would be greatly appreciated!
  dT       d2T
 ---- = A  ----
  dt       dx2
Initial Condition
T=1 at t=0 for all x
First Boundary Condition
T=1 at x=0 for t>0
Second Boundary Condition
  dT
 ---- = KT    at x=1  for t>0
  dx

Gabe


Palmerston North, New Zealand - Wednesday, November 30, 2005 at 22:24:56 (EST)
Reply to Jenia: Are you taking the limit from above or below? From above, that is x > 0, then

   lim
  x → 0+
     x
            =
  |sin(x)|
   lim
  x → 0
    x
        
  sin(x)

but from below with x < 0

   lim
  x → 0-
     x
            =
  |sin(x)|
   lim
  x → 0
     x
         
  -sin(x)


Karl <Click to Send Email to Karl>
USA - Wednesday, November 30, 2005 at 17:58:17 (EST)


i'm looking for the limit of the following function
lim(x--->0) x/abs(sin(x))
i think it should be 1 since its very similar to the function
x/sin(x)
but i'm not sure and don't know how to prove it.
Jenia


Jerusalem, Israel - Wednesday, November 30, 2005 at 15:24:25 (EST)
Reply to Lisa: Just listing a function is not stating a problem. You have to say what the problem asks you to do with it -- differentiate, integrate -- what?

Your function can be simplified using identities and algebra to make it much easier to differentiate or integrate. First identity is ln(√b) = (1/2)ln(b). So

  ln(√x² - 1)  =
  1
   
  2
 ln(x2 - 1)

You can also factor x2 - 1 = (x + 1)(x - 1). And then you can apply ln(ab) = ln(a) + ln(b). So

  1
   
  2
 ln(x2 - 1)  =
  1
   
  2
 ln(x + 1)  +
  1
   
  2
 ln(x - 1)

I'll leave it to you to integrate or differentiate this on your own because with the function in this form, either operation is easy.
Karl <Click to Send Email to Karl>
USA - Wednesday, November 30, 2005 at 08:56:14 (EST)


Hello, I have a really bad teacher who really has a hard time explaining anything to us. I am at a loss with this problem: f(x)=ln√ x²-1
Lisa


Georgetown, KY USA - Tuesday, November 29, 2005 at 21:37:59 (EST)
Reply to Laura: You didn't say what you wanted done with the expression, x3(5x2 + 1)-2/3. If you want the derivative of this, you can see that this function is the product, u(x)v(x), where u(x) = x3 and v(x) = (5x2 + 1)-2/3. So you would apply the product rule. To find the derivative of the second factor, v(x), observe that v(x) = g(h(x)) where g(h) = h-2/3 and h(x) = 5x2 + 1. So you would apply the chain rule.

If you want to find the integral

  
 x3(5x2 + 1)-2/3 dx

then substitute u = 5x2 + 1. Hence du = 10x dx or equivalently du/10 = x dx. Also you have (u-1)/5 = x2. Making the substitution, the integral becomes: