Karl's Calculus Forum: Dec 2002

Archived Forum Discussion for December 2002

Reply to Steve Barrow: Suppose I asked you to find out what fraction of an foot is left over measuring the distance between the tip of the forward spike on the Statue of Liberty's crown and the tip of the minute hand on Big Ben at precisely noon London time. And suppose I gave you a ruler to measure it with. Now let's suppose that you had some method for measuring along the most direct possible path where you could lay the ruler down again and again, marking off the feet, despite having to do it over the Atlantic Ocean. Let's just say you could mark off the lengths of the ruler perfectly. How accurate does that ruler have to be to make your measurement worth anything? Well the distance between those two objects is, according to my almanac, about 18 million feet. So clearly, if you want to know the fraction of a foot left over, your ruler would have to be accurate to better than one part in 18 million.

So where am I going with this? Your attempt to find the sine of 10⁴⁰ is very much the same problem. You need a ruler that is 2p long. You want to measure 10⁴⁰ with it and find what fraction of your ruler length is left over. Guess what? Your ruler needs to be accurate to better than 1 part in 10⁴⁰. That means you need to know p out to more than 40 places. And you need to be able to do arithmetic accurate to that many places as well. If your arithmetic is any less accurate, then you are not using the full accuracy of your approximation of p. So there is no short-cut around having to do arithmetic out to extreme precision in order to find the remainder of 10⁴⁰ divided by 2p. And you need to know that remainder in order to find the sine of 10⁴⁰.
Karl <Click to Send Email to Karl>
USA - Monday, December 16, 2002 at 19:47:36 (EST)

I have been struggling for ages trying to find an accurate method to finding and proving the value for sin 1 E40. I have had no luck so far, please help me! thanks Steve Barrow
Steve Barrow

Devon, SW UK - Sunday, December 15, 2002 at 15:27:06 (EST)

brittany, On the first one, you need to take the first deriviative of each equation, set it equal to zero, find the critical points and based on what the sign of the derivitive is between the critical points, decide whether its a max or min.

I'll do part a. to get you started
y = 10 - 5x - x^2

y' = -5 - 2x

now we set y' = 0,  and solve for x

0 = -5 - 2x

x = -(5/2)

now the easiest way (for me at least) to anyalize this is to make a number line like this...

      |____________________(-(5/2))___________________|

now we test at points to the left and right of the point say...(-6/2) and 0,
for -6/2, y' is positive, and for 0 y' is negative, so we put that on the number line

        ++++++++++++++++++         ------------------
      |____________________(-(5/2))___________________|

since there is only 1 critical point, we can see that the slope goes
from positive to negative here, this will be a maximum, in fact an
absolute one, the function is a negative parabola.
to find the maximum value itself, plug -5/2 into the original
equation.

for part b. you need to take the second derivative y'', and anylize it
in a similar fashion, but the points of inflection will tell you where
the concavity changes instead of the slope.

hope that was helpful!


Justin







 USA - Sunday, December 15, 2002 at 02:00:39 (EST)


Karl,
I figured out the last two questions but am still stuck on the simpler first two.

For each of the following, find the extreme value(s) and determine whether each is a minimum or a maximum.
a.  y = 10 – 5x + x²		b.  y = 12x – 6x² + x³

2.  For each of the following functions, determine if the function is concave or convex, or both.  If it changes, state the domain over which it is convex and the domain for which it is concave.
a.  y = x² +3x - 3		b.   y = 12x – 6x² + x³

Brittany
USA - Saturday, December 14, 2002 at 15:09:31 (EST)


Sorry, I'm going to format my question better..

Please help Karl. I have 3 questions. They involve optimumization which I'm not too strong in:

1.For each of the following, find the extreme value(s) and determine whether each is a minimum or a maximum.
a.  y = 10 – 5x + x²		b.  y = 12x – 6x² + x³
2.For each of the following functions, determine if the function is concave or convex, or both.  If it changes, state the domain over which it is convex and the domain for which it is concave.
a.  y = x² +3x - 3		b.   y = 12x – 6x² + x³

3.For the total cost function given by C(Q) = Q³ – 8Q² + 30Q + 5, find the firm’s marginal cost (MC) and average cost (AC) functions.(for MC i would just find the deriviative right..?? For AC i would just divide by Q right..??)

4.  If the firm from question 4 firm faces the following demand function:   Q = 40 – 2P
a.  Find the profit function for the firm.
b.  Use the first order conditions to find the relative extremum.
c.  Check the second-order conditions and determine what the profit maximizing level of profit is.
d.  What is the maximum profit?

Thanks for any help...


Brittany


Brittany

 USA - Saturday, December 14, 2002 at 08:56:29 (EST)


Please help Karl. I have 3 questions. They involve optimumization which I'm not too strong in:

1.For each of the following, find the extreme value(s) and determine whether each is a minimum or a maximum.
a.  y = 10 – 5x + x2		b.  y = 12x – 6x2 + x3

2.For each of the following functions, determine if the function is concave or convex, or both.  If it changes, state the domain over which it is convex and the domain for which it is concave.
a.  y = x2 +3x - 3		b.   y = 12x – 6x2 + x3

3.For the total cost function given by C(Q) = Q3 – 8Q2 + 30Q + 5, find the firm’s marginal cost (MC) and average cost (AC) functions.(for MC i would just find the deriviative right..?? For AC i would just divide by Q right..??)

4.  If the firm from question 4 firm faces the following demand function:   Q = 40 – 2P
a.  Find the profit function for the firm.
b.  Use the first order conditions to find the relative extremum.
c.  Check the second-order conditions and determine what the profit maximizing level of profit is.
d.  What is the maximum profit?

Thanks for any help...

Brittany
USA - Saturday, December 14, 2002 at 08:11:01 (EST)

Reply to Candice: Justin is right that you can't solve for dy/dx without there being an equal sign, so to demonstrate how to solve these, I will add the equal and solve

xy² + xy^1/2 + 2  =  0

for dy/dx.

Start with the first product, xy². Since it is a product, you use the product rule. The derivative of x is 1. The derivative of y² is 2y dy/dx. Notice that because y is the dependent variable (which it usually is), that means that y is a function of x. Since we are taking derivatives with respect to x, we have to multply the 2y by the derivative of y(x), which is in accordance with the chain rule. So the derivative of the first product is y² + (2xy dy/dx). You do the second product the same way. The derivative of y^1/2 is (1/2)y^-1/2 dy/dx. So the whole equation's derivative is:

   y² + (2xy dy/dx) + y^1/2 + ((1/2)xy^-1/2 dy/dx)  =  0

Put all the stuff that has a dy/dx in it on one side of the equation:

   y² + y^1/2  =  -(2xy dy/dx) - ((1/2)xy^-1/2 dy/dx)

Now factor the dy/dx out of the right-hand side, and divide both sides of the equation by the remaining factor. That will leave you with dy/dx all alone on the right, and you'll be done.

Hope this saves your life :-)
Karl <Click to Send Email to Karl>
USA - Thursday, December 12, 2002 at 19:39:52 (EST)

Reply to Jennifer Par: I can't do sketches for you in this medium, but I can give you tips. The first area is bounded by:

n) x+y=1, x+y=5, y=2x+1, y=2x+6              answer=20/3

The first two lines are parallel to each other. Also the third and fourth lines are parallel to each other. So this area is a parallelogram. It has four vertices. Find them by finding the intersection points between the first line and each of the third and fourth (that gives you two of the vertices) and then by finding the intersection points between the second line and each of the third and fourth (that gives you the other two vertices). Put the four points in order according to their x coordinates. Now integrate the difference of the fourth minus the first from the first point to the second. Integrate the second minus the first from the second point to the third. Integrate the second minus the third from the third point to the fourth. Add all those areas together to get your answer. When you draw the plot of all the lines it will be clearer what to do.

The second is:

t) y=3sinx, y=sin3x from 0 to pie                 answer=16/3

Integrate (3 sin(x) - sin(3x))dx from zero to p. Remember that the integral of 3 sin(x) is -3 cos(x) + C, and the integral of sin(3x) is -(1/3)cos(3x) + C.

The last one is:

3c) y=x+1, y²=2x+6                        answer=18

The second function is the same as y = ±sqrt(2x+6). But to start, replace the y in the second equation with the x+1 from the first equation:

  (x+1)²  =  2x + 6

This you can solve using the quadratic formula to find x value of the two intersection points. Those will be the limits of your integration. Now integrate

   ____
 (Ö2x-6  -  x+1) dx

from the lower limit to the upper limit. The result is your answer.
Karl <Click to Send Email to Karl>
USA - Thursday, December 12, 2002 at 19:13:29 (EST)

Hi Karl I desperatley need your help. I could not figure out the following questions. 1.Find the area of the region between the given curves. Include a sketch of the region

n) x+y=1, x+y=5, y=2x+1, y=2x+6              answer=20/3

t) y=3sinx, y=sin3x from 0 to pie                 answer=16/3

3c) y=x+1, y2 (squared)=2x+6                        answer=18

thank you so much i would really appreciate it if you could help me
Jennifer Par

Fresno, CA USA - Thursday, December 12, 2002 at 11:46:33 (EST)

candice, yeah implicit differentiation is hard when there is no equality there. are you sure thats the whole problem, not of the form f(x,y) = g(x,y)?? I dont know of any way to do implicit differentiation without this form. let me know and i can help
Justin

USA - Wednesday, December 11, 2002 at 19:29:24 (EST)

we have a huge take home test and i need help with a problem. i can't do implicit differentiation to save my life and there are two problems that are keeping me stuck. they look like this: Find dy/dx by implicit differentiation. xy^2 + xy^1/2 + 2 please help. candice
Candice

USA - Tuesday, December 10, 2002 at 22:19:03 (EST)

Dear D. Simmons, Solving this problem is somewhat simple. First you have to graph the equation and the boundaries. The area that you are evalating is between f(x)= 1/3x(squared) and y=x in the first quadrant. Then if you are integrating in terms of x try to visualize what that segment would look like as it spins around the x-axis, creating a 3-d image. If there is no hole through the middle, it is a disk, wheras vice versa would be a washer. Since this segment forms a disk you should integrate pie, times, f(x)squared, times dx using the interval between the intersections of f(x) and y=x if you are solving in respect to y, the segment is now spinning around the y-axis,let y = 1/3x(squared) = f(x) and y = x = f(x). Now you can calculate the volume by integrating pie times(f(x)squared - g(x)squared)dy between the interval of the y-values at the intersection of the two equations. note: these two formulas are universal for all problems, but be careful because when solving with respect to y f(x) will be the line on the right side of the segment and g(x) will be the line on the left good luck w/your problem and if you need any more help or have questions, feel free to ask -claire ps- the two equations that are used to find the volume are just simply a calculation of area where the equation of the functions act as the radius
claire
CO USA - Monday, December 09, 2002 at 19:31:30 (EST)

Sarah, implicit differentiation is the right approach, but this problem is even messier than the last, differentiate both sides with respect to x to get:

3*((3x+y)^2)*(3 + dy/dx) = 2*(x - 7y + 1)*(1-7*dy/dx)

Now its a matter of collecting terms and solving for dy/dx. Justin
Justin

USA - Monday, December 09, 2002 at 16:37:40 (EST)

How do I solve this.....the instructions are Use the method of disks and washers and the method of cylindrical shells to give one integral with respect to x and one with respect to y, each of which equals the volume generated by rotating the region about the indicated axis. Then evaluate both inegrals. I got the right answer by using the cylinddrical shell method with regards to x, but can't seem to set up in regard to y using the method of disks and washers. Here's the question: The region bounded by y= 1/3(x squared), y=x, axis: x=0 Thank You in advance D.Simmons
D. Simmons

USA - Monday, December 09, 2002 at 15:05:07 (EST)

Reply to Chris: To integrate your function, first substitute u = sin(x) and du = cos(x) dx. This gives:

   du
       
 1 + u²

That can be integrated by substituting v = tan(u) and dv = (1 + tan²(u)) du. See if you can take it from there.
Karl <Click to Send Email to Karl>
USA - Monday, December 09, 2002 at 12:54:44 (EST)

Reply to SA: This is the problem to minimize the cost of the box. Your constraint is that

   V  =  LWH  =  12

where L is length, W is width, and H is height. You are probably familiar only with constraint equations that have two independent variables. This one has three, which adds a complication that might be new to you. There are two ways you can do it. One is by Lagrange multipliers, and if you have covered that topic, that is the way to go. If not, then you can do it with partial derivatives. Your cost equation is

   C  =  4LW  +  6LH  +  4WH

First use the constraint equation to solve for H in terms of L and W. Substitute that expression into the cost equation for H.

Now find the partial derivatives, ¶C/¶L and ¶C/¶W. That will give you two separate equations. Set both partials to zero and solve for L and W simultaneously. Back-substitute to find H.
Karl <Click to Send Email to Karl>
USA - Monday, December 09, 2002 at 12:49:23 (EST)

Reply to Rachel: If the object is propelled upward, then its initial velocity is v(0). If it starts from 10 feet up, then its equation of motion is:

            1
   x(t)  =    at²  +  v(0)t  +  10 ft
            2

You know that a = -32 ft/sec². Put in 10 seconds for t and zero for x(t) (because the ground is at a height of zero). Now solve for v(0).
Karl <Click to Send Email to Karl>
USA - Monday, December 09, 2002 at 12:39:06 (EST)

Reply to Rachel: You have acceleration of

   a(t) = 6 + 12t

Integrating twice you find:

   v(t)  =  6t + 6t² + C

   x(t)  =  3t² + 2t³ + Ct + D

where C and D are yet-to-be-determined constants. The problem gives you initial conditions of v(0) = 0, and x(0) = 1. Put the first initial condition into the v(t) equation to solve for C. Then put that solution for C along with the initial condition for x(0) into the x(t) equation to solve for D.
Karl <Click to Send Email to Karl>
USA - Monday, December 09, 2002 at 12:32:12 (EST)

     f(x)= integral of( cos(x)
                        ------
                        1+sin^2x) dx
thanks for all help


Chris







LA, CA USA - Sunday, December 08, 2002 at 20:25:40 (EST)

Thanks Justin, I have another one I am stuck on =( I need to find the derivative of this one. I tried using implicit differentiation, is this the right approach? Thanks for any help =) !!

(3x+y)^3=(x-7y+1)^2

Sarah

USA - Sunday, December 08, 2002 at 19:36:34 (EST)

Hi, I was wondering if anyone could help me with an optimization problem: A rectuangular box with no top is to be constructed to have a volume V=12ft^3. The cost per square foot of the material to be used is $4 for the bottom, $3 for two of the opposite sides, and $2 for the remaining pair of opposite sides. Find the dimensions of the box that will minimize the cost. I could do this problem if there wasn't a restriction on the voulme, but the restriction on the volume is making this a harder problem to do. Any pointers would be greatly appreciated. Thanks. SA
S A

USA - Sunday, December 08, 2002 at 10:21:20 (EST)

Sarah, for that derivitive you must take the natural log of both sides

ln (y) = ln (x^tanx)

ln (y) = tan(x)*ln (x)

and raise both sides to the e to get:

y = e^(tan(x)*ln(x))

which can be differentiated using the chain rule and the product rule, its a mess but it works!
Justin

Dallas, TX USA - Saturday, December 07, 2002 at 19:31:47 (EST)

How do you take the derivative of x raised to a a trig function? For example:

y = x^(tanx)

Thanks for any help!
Sarah

USA - Saturday, December 07, 2002 at 16:42:55 (EST)

Find the largest possible area for a rectangle with base on the x-axis and the upper vertices on the curve y=2-x².
Rachel

Hinesville, GA USA - Saturday, December 07, 2002 at 13:26:44 (EST)

Compute the tangent line for y equals the square root of (x-1) at x = 5.
Rachel

Hinesville, GA USA - Saturday, December 07, 2002 at 13:19:20 (EST)

An object is propelled vertically upward from a height of 10 feet. Determine the initial velocity so that this object reaches the ground after exactly 10 seconds.
Rachel

Hinesville, GA USA - Saturday, December 07, 2002 at 13:12:13 (EST)

Given the accelerarion a(t) = 6+12t, the initial velocity equals zero and the initial position equals 1. Compute the position x(t).
Rachel

Hinesville, Ga USA - Saturday, December 07, 2002 at 13:06:25 (EST)

Reply to Mike: First take the log of both sides:

   ln(f(x))  =  ln(x^(1/cos(x))  =  ln(x) / cos(x)

Now take the derivative, using the chain rule to do the left-hand side and the quotient rule to do the right-hand side:

   f'(x)      cos(x)/x  +  ln(x)sin(x)
          =                           
   f(x)              cos²(x)

You know what f(x) is from the original equation given in the problem. Multiply through by it, and you're left with the solution for f'(x).
Karl <Click to Send Email to Karl>
USA - Thursday, December 05, 2002 at 21:38:32 (EST)

Reply to Ashwin S: Applying L'Hopital's rule to the original limit, you find that

           f'(x)
    lim           =  1
   x -> 0   2x

So clearly f'(0) = 0. Make sure you see why.

For x very close to zero, you have f(x) » f(0) + x f'(0). But you already figured out that f(0) = 0. You are looking for the limit

           f(x)           x f'(0)
    lim          =  lim            =  lim   f'(0)
   x -> 0    x     x -> 0    x       x -> 0

You should be able to get it from there.
Karl <Click to Send Email to Karl>
USA - Thursday, December 05, 2002 at 21:17:12 (EST)

Reply to Mike: Then integrating once you have

   f'(x)  =  (1/3)x³  +  (2/3)x^3/2 + C

where C is some constant. Integrating again

   f(x)  =  (1/12)x⁴  +  (4/15)x^5/2 + Cx + D

where D is some other constant. But the problem tells you that f'(0) = 1. Put that into the equation for f'(x) to solve for C. Once you know C, put it and f(0) = 1 into the second equation to solve for D.
Karl <Click to Send Email to Karl>
USA - Thursday, December 05, 2002 at 21:06:44 (EST)

I am stuck on the following problem:

If lim  f(x)
   x->0 ---- = 1,
         x^{2
find:
a. lim f(x)
   x->0
b. lim  f(x)
   x->0 ----
         x}

I get the answer to (a) as 0, but then what can I do to the limit in (b)? I can't factorise because the definition of f(x) is not given. Does this have something to do with the Sandwich Theorem? Any advice would be appreciated. Thanks, Ashwin.
Ashwin S

Bombay, MAH India - Wednesday, December 04, 2002 at 10:43:07 (EST)


I need some help w/ these two problems.

Use logarithmic differentiation to find the derivative of
f(x) = x^(1/cos(x))

Find f, if f''(x) = x^2 + sqrt(x), and f'(0) = f(0) = 1.


Thanks,
Mike

Mike
NYC, NY USA - Tuesday, December 03, 2002 at 19:25:55 (EST)

Reply to drea: The problem does not ask you to maximize the number of items sold. It asks you to maximize profit. Profit is $5 times the number sold minus what is spent on advertising:

   p  =  10000(1 - e^-0.001n) - n

   dp
       =  10e^-0.001n  -  1
   dn

Now find the n that makes the derivative (second equation above) equal to zero.
Karl <Click to Send Email to Karl>
USA - Tuesday, December 03, 2002 at 18:42:33 (EST)

5. A company makes a profit of $ 5 on each items of its product it sells. If it spend n dollars per week on advertising, then the number of items per week it sells is given by x = 2000(1- e^-kn) where k= 0.001. Find the value of n that maximize the profit. I have brought this down to the first derivative but cannot seem to solve for zero did i bring this problem down wrong? First derivative = x'=2e^^-.001n? Help?!?!
drea

USA - Tuesday, December 03, 2002 at 15:32:54 (EST)

hi everyone (first timer) can someone please help me figure out how I can turn $3 per foot into meter, I think it's $9.2 per meter but I am not sure. And also how do you minimize a cost. thank you!
Shelly

Dixon, CA USA - Monday, December 02, 2002 at 16:45:48 (EST)

Trying to prove:

If alpha:[a,b]--->Rⁿ, beta:[c,d]--->Rⁿ are smooth, one to one paths such that alpha(a)=beta(c) and alpha(b)=beta(d).  Show that phi=beta^-1 composed alpha is C¹.

Rose







 USA - Monday, December 02, 2002 at 15:07:05 (EST)

Reply to CJ HENRY: You have to use a combination of the chain rule and the product rule to do the left-hand side. Note that the left-hand side is a composite, f(g(x,y)), where f(g) = e^g, and g(x,y) = xy. So the derivative, according to the chain rule, will be f'(g(x,y))g'(x,y), where g'(x,y) is the implicit derivative of xy. But to find that, you use the product rule: implicit derivative of xy is xy' + y. Putting it all together, you have the implicit derivative of the equation, e^xy = sin(x):

   e^xy (xy' + y)  =  cos(x)

Now simply solve for y'
Karl <Click to Send Email to Karl>
USA - Sunday, December 01, 2002 at 14:19:13 (EST)


 find dy/dx using implicit differentiation.

e^{xy  = sinx}

CJ HENRY

HUNTSVILLE, AL USA - Sunday, December 01, 2002 at 11:40:29 (EST)

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