Karl's Calculus Forum: July-December 2006

Archived Forum Discussion for July-December 2006

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) = x^ln(x):

Use the identity:

   b^x = e^{x ln(b)}

So what you really need to take the derivative of is

   e^{ln(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)= x^ln(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 16^3x = (2⁴)^3x. Remember the rule for taking a power of a power -- you multiply the exponents. So (2⁴)^3x = 2^12x.
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)

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