Karl's Calculus Forum: Nov 2002

Archived Forum Discussion for November 2002

Reply to Brittany/Michael J./Dr. Gonzo: Who'd have thought that there would be three people sharing the same cable-modem who were all taking advanced calculus. What are the odds -- it boggles the mind.

To do this problem, find the two partial derivatives of U:

   ¶U
       =  2(T^-4/5)(S^4/5)
   ¶T

   ¶U
       =  8(T^1/5)(S^-1/5)
   ¶S

The total derivative is:

          ¶U        ¶U
   dU  =     dT  +     dS
          ¶T        ¶S

You have to know what T and S are to find the numerical value that is the solution. That's because T and S are involved in the expressions for the partials you need to find dU.

Once you have the partials, substitute dT with the delta given in the problem for T and substitute dS with the delta given in the problem for S. Put that along with the values for T and S into the expression for dU, and the resulting dU is your answer.
Karl <Click to Send Email to Karl>
USA - Saturday, November 30, 2002 at 21:03:59 (EST)


I left out that U(T,S)

Brittany
USA - Friday, November 29, 2002 at 18:37:57 (EST)


The function is:

U=10T^(1/5)*S^(4/5)

The directions for this problem state:
Totally differentiate the utility function in order to determine how your utility will change if you consume three more units of T and four less units of S.

Thanks

Brittany
USA - Friday, November 29, 2002 at 18:32:26 (EST)

Thanks...
Michael J.
USA - Friday, November 29, 2002 at 17:48:19 (EST)

Reply to Michael J. When you take a partial derivative with respect to a particular variable, you have to treat all other variables as if they were constants. That is how a partial derivative is defined. So for

  r(w,x,y,z)  =  w - (2x + 3y + 4z + 6)/(2x)

use the rules for taking derivatives (you will need to use the quotient rule to find the partial with respect to x). Here are your partials for r(w,x,y,z):

  ¶r
      =  1
  ¶w

  ¶r       (2x)(2) - (2x + 3y + 4z + 6)(2)
      =  -                                
  ¶x                   (2x)²

  ¶r        3
      =  -   
  ¶y       2x

  ¶r        4
      =  -   
  ¶z       2x

You might want to review your textbook on how to take partials.
Karl <Click to Send Email to Karl>
USA - Friday, November 29, 2002 at 17:16:44 (EST)


With regards to the previous post, what would I do with
r(w,x,y,z)  =  w - (2x + 3y + 4z + 6)/(2x)??

w=1, y=3, z=4, but what about x...?? Can I simplify before i take the partial derivative..??

Michael J.
USA - Friday, November 29, 2002 at 15:43:52 (EST)

Reply to Michael J. The Jacobian determinant (the way I learned it) is applicable if you have functions of the variables in question. In your case you have four variables, so you would need four functions. I'll show you how it's done with just three variables and three functions (I'm simplifying the example matrix to 3x3 because the editor won't allow me to type a 4x4): If you had the three functions, p(x,y,z), q(x,y,z), and r(x,y,z), then the matrix is:

   ¶p  ¶p  ¶p
             
   ¶x  ¶y  ¶z

   ¶q  ¶q  ¶q
             
   ¶x  ¶y  ¶z

   ¶r  ¶r  ¶r
             
   ¶x  ¶y  ¶z

You would then find the determinant of this matrix by the usual method of finding determinants.

Since you have four variables, you would need four functions. But you haven't provided functions here, just equations (three of which are linear). You could do something like this:

   2w + x - z = 1

   becomes    p(w,x,y,z)  =  2w + x - z - 1

In which case your partials for the first row are:

   ¶p           ¶p
       =  2         =  1
   ¶w           ¶x


   ¶p           ¶p
       =  0         =  -1
   ¶y           ¶z

which would form the first row of your matrix. Do similar for each of the other equation:

  w + x = y + z

  becomes  q(w,x,y,z)  =  w + x - y - z

  w = (2x + 3y + 4z + 6)/(2x)

  becomes  r(w,x,y,z)  =  w - (2x + 3y + 4z + 6)/(2x)

  x - z = 2w

  becomes  s(w,x,y,z)  =  -2w + x - z

Find all the partial derivatives of each of these to complete your 4x4 matrix. Then find the determinant.
Karl <Click to Send Email to Karl>
USA - Friday, November 29, 2002 at 15:28:52 (EST)


I need to find the Jacobian Determinant for a system of 4 equations but am having trouble setting up the matrix:

2w+x-z=1                  w+x=y+z
w=(2x+3y+4z+6)/2x         x-z=2w

Michael J.
USA - Wednesday, November 27, 2002 at 21:07:56 (EST)

Oh, i figured it out. Thanks anyway...
Dr. Gonzo
USA - Wednesday, November 27, 2002 at 21:02:55 (EST)


Let me clarify my previous post: I need to differentiate the following functions with respect to x. Would your previous comments still apply??

     f= (x/y) + (2w/x)

     g= 2x⁵ln(5x/1-x)

     h= 12+e^{Ax^Alphay^Beta}

Thanks




Dr. Gonzo

 USA - Wednesday, November 27, 2002 at 14:08:39 (EST)


I need to find the Jacobian Determinant for a system of 4 equations but am having trouble setting up the matrix:

2w+x-z=1                  w+x=y+z
w=(2x+3y+4z+6)/2x             x-z=2w

Michael J.
USA - Wednesday, November 27, 2002 at 13:52:57 (EST)

Reply to Dr. Gonzo: Since you are asking for dy/dx in each case, I shall assume that f, and h are constants, because if they are functions of x and y, then the answers would involve partial derivatives of such functions. I shall also assume that y is a function of x.

The first and third problems involve implicit differentiation. On the first one, use that together with the quotient rule:

   0  =  (y - xy')/y²  -  w/x²

where w is assumed constant. Finish by solving for y'.

The second one is made easier using the log identity:

   ln(5x/1-x)  =  ln(5x) - ln(1-x)

I'll let you do the rest of that one (you didn't include any y in this one, so I expect the problem is asking for dg/dx). Use the product rule to finish this one.

The last one requires use of the chain rule and product rule.

   0  =  e^abxy ab(xy' + y)

where a and b are assumed constant. Finish by solving for y'.
Karl <Click to Send Email to Karl>
USA - Wednesday, November 27, 2002 at 13:03:29 (EST)


Find dy/dx of the following:

     f= x/y + 2w/x

     g= 2x⁵ ln(5x/1-x)

     h= 12+e^{Ax^Alphay^Beta}

Dr. Gonzo
USA - Wednesday, November 27, 2002 at 10:31:59 (EST)

Reply to Scott: You need to use the binomial formula to expand (t+h)³ and (t+h)².

   (t + h)³  =  t³ + 3t²h + 3th² + h³

   (t + h)²  =  t² + 2th + h²

Put those expression in for your powers of (t+h) and see what happens (don't forget to distribute the original coefficients over all the terms).

One more thing. You don't set h=t at the end. You take the limit as h goes to zero. After you've canceled all the h's from numerator and denominator, what are you left with as h approaches zero?
Karl <Click to Send Email to Karl>
USA - Tuesday, November 26, 2002 at 18:53:05 (EST)

Ok, I am heading back to college after quite a few years to finish my Mechanical Engineering degree. As preperation for this, I was going through a whole bunch of the old classes I had taken before. It really isn't the advanced concepts that escape me, and not even concepts at all, I am just having an issue with doing derivitives out long, as I have done them using short tricks for so long, I am not sure what is wrong here. Let me show you: Let's find the derivative of this to start:

   f(t) = 3t³ + 4t² - 6

OK, doing it all shorthand, I can easily tell that the answer is:

   f'(t) = 9t² + 8t

But if I do it out long, it goes like this:

   f'(t) = ((3(t+h)³ + 4(t+h)² - 6)-(3t³ + 4t² - 6))/h

Now, distribute this out:

f'(t) = (3t³ + 3h³ + 4t² + 4h² - 6 - 3t³ - 4t² + 6)/h

A number of these have like terms, so leave us with:

f'(t) = (3h³ + 4h²)/h

A little quick and simple division leaves:

f'(t) = 3h² + 4h

And because we are looking for like terms and h happens to be arbitrary, we will assume h=t , giving us a final answer of:

f'(t) = 3t² + 4t

Huh? This answer is not at all the correct answer. Does anyone see where I went wrong?
Scott
Colorado Springs, CO USA - Tuesday, November 26, 2002 at 10:55:53 (EST)

Reply to Lynn: Find the second derivative of this function:

   f'(x)  =  3x² - 6x - 9

   f"(x)  =  6x - 6

Now find which values of x make f"(x) negative.
Karl <Click to Send Email to Karl>
USA - Friday, November 22, 2002 at 23:24:54 (EST)

Reply to Fran: You have

  R'(n)  =  -12n³ + 150n² - 522n + 540

You need to solve for R'(n) = 0. First factor out 6 from all terms:

   -2n³ + 25n² - 87n + 90  =  0

You can factor this thing, but I'll make it easier for you by telling you that one of the roots is n = 2. Use polynomial long division to divide n - 2 out of the polynomial above. Then you can use the quadratic formula to find the other two roots.

Each of the roots is a candidate for the n that optimizes the original revenue equation. Try each root and see which maximizes revenue.
Karl <Click to Send Email to Karl>
USA - Friday, November 22, 2002 at 23:21:12 (EST)

Reply to mjy: Applying the quadratic formula to

   4x² - 32x + 45  =  0

you find

                __________
   x  =  (32 ± Ö1024 - 720) / 8

   x  =  4  ±  2.1794499

Use the second derivative test to see which of these solutions is a max and which is a min. The second derivative of volume is

   V"(x)  =  3(8x - 32)

Where V" is negative, a critical point will be a max. When you find the x solution for the max, simply substitute it back into the volume formula,

   V(x)  =  (15 - 2x)(9 - 2x)(x)

to get the max volume.
Karl <Click to Send Email to Karl>
USA - Friday, November 22, 2002 at 08:08:06 (EST)

Help. I am not able to understand this one. 'determine where the graph of f(x) = x^3-3x^2-9x+1 is concave down'? I think it is x<1 but I am unsure. Your help is appreciated!
Lynn
USA - Thursday, November 21, 2002 at 18:59:09 (EST)

I would appreciate your help in figuring out this problem: Personnel Manager estimates that if she hires temp salespeople the total net revenu derived from thier efforts will be: R(n) = -3n^4 + 50n^3 = 261n^2 + 540n hundred dollars for 0<= n <= 10 How many salespeople would need to be hired to maximize the total net revenue. This one has me stumped. Thank you in advance for your help!
Fran
USA - Thursday, November 21, 2002 at 18:00:23 (EST)

Hi I need to find the max volume of a open box. Starting with rectangle with sides 15' by 9'. I get the first derivative as 3(4x^2 - 32x + 45) I use the quadratic, but then cannot get algebra for substituting answer for x in (15-2x)(9-2x)(x)
mjy
W, PA USA - Wednesday, November 20, 2002 at 20:58:51 (EST)

Reply to Frank Caron: You apply the product rule to find the derivative of the first term. The second term is easy.

   f'(x)  =  sin(x)  +  x cos(x)  -  sin(x)  =  x cos(x)

Now set this to zero to find the critical points:

   0  =  x cos(x)

A product can be zero only when at least one of the factors is zero. So you have a critical point at x = 0 from the first factor and critical points at x = ±p/2 from the second factor. Use the second derivative test to see which of those is a max and which is a min. Then find f(x) for each of the critical points and compare them with f(x) at the endpoints of the interval in question -- remember your absolute max or min can be at an endpoint. See if you can take it from there.
Karl <Click to Send Email to Karl>
USA - Wednesday, November 20, 2002 at 20:01:51 (EST)

Reply to Tom: The first one is definitely a chain rule problem. Since it is a composite of a composite, you have to apply the chain rule twice. The chain rule requires (for a composite of a composite) if:

  f(x)  =  u(v(w(x)))

then

   f'(x)  =  u'(v(w(x)))v'(w(x))w'(x)

In your case, w(x) = tan(x), v(w) = ln(w), and u(v) = u². Your derivatives are: w'(x) = sec²(x), v'(w) = 1/w, and u'(v) = 2v. See if you can put it all together now.

For second derivative of cos(2x). Again apply the chain rule. The derivative of 2x is simply 2. So the first derivative of this function is -2 sin(2x), and the derivative of that is -4 cos(2x).
Karl <Click to Send Email to Karl>
USA - Wednesday, November 20, 2002 at 19:54:05 (EST)

Reply to dr0dlr: First one, use the product rule:

  g'(s)  =  2s cos(s)  -  s² sin(s)

To take the second derivative, use the product rule again (twice, once on each summand):

  g"(x)  =  2cos(s)  -  2s sin(s)  -  2s sin(s)  -  s² cos(s)

which you can simplify.

For the second, presumably y is a function of x. Then you have

   x^1/2  +  y^1/2  =  1

applying the power rule to both terms and the chain rule to the second term:

   (1/2)x^-1/2  +  (1/2)y^-1/2 y'  =  0

You can solve for y' in terms of x and y from that.
Karl <Click to Send Email to Karl>
USA - Wednesday, November 20, 2002 at 19:45:36 (EST)

Hey Karl, you got a great site here, it is helping me keep my calculus mark up. Man, you answer just about everything and I cherish your examples. You are an excellent teacher. Right now, I am doing Grade 13 Calculus and I need some help. We are currently doing Derivatives of Trig Functions and their applications. Since you don't have a application sections for Trig Functions, I'll write out the problem I am having trouble with: 1) Find the local max/min of the function: f(x) = x[sin (x)] + cos (x), with the domain being -pi less than or equal to x less than or equal to +pi
Frank Caron
Mississauga, Ontario Canada - Wednesday, November 20, 2002 at 16:57:57 (EST)

I need some asssitance with the following.

1. Differentiate the function y = (ln(tan(x)))^2

2. Find the first and second derivative of the function y = cos(2Ø).

Thanks,
Tom

Tom
USA - Tuesday, November 19, 2002 at 21:26:11 (EST)

Could you please help me w/ these 2 problems? Thanks in advance.

Find first and second derivative.

g(s) = s^2cos(s)


Differentiate implicitly.

sqrt(x) + sqrt(y) = 1


Thanks Again.

dr0dlr USA - Monday, November 18, 2002 at 18:15:51 (EST)

Reply to Shannon: The volume of water is the crossectional area of the water times the length of the trough. If the water is h m deep (be sure to convert cm to meters), then the crossectional area of the equilateral triangle is

   A  =  (sqrt(3)/3)h²

which you get from the geometry of an equalateral triangle. And volume is given by

   V  =  L (sqrt(3)/3)h²

where L is the length of the trough. Remember that h and V are both functions of time, t. So take the derivative of both sides with respect to t (using implicit differentiation on the right), and you find:

   dV                      dh
       =  2L (sqrt(3)/3) h   
   dt                      dt

The problem gives you dV/dt, L, and h. You need to solve for dh/dt.
Karl <Click to Send Email to Karl>
USA - Monday, November 18, 2002 at 12:48:51 (EST)

       ax+1^1/3-1
lim =  --------
x->0      x

I don't know how to get rid of the cubed root! You can only solve with limits...help please! I tried everyone that knows math!
Omar oakville, ont canada - Sunday, November 17, 2002 at 20:50:17 (EST)

Problem: A cross-section of water trough is an equilateral triangle with the top edge horizontal. If the trough is 5m long and 25cm deep and the water is flowing at a rate of 0.25m³/min, how fast is the water level rising when the water is 10cm deep at the deepest point????
Shannon
ON Canada - Sunday, November 17, 2002 at 12:51:04 (EST)

Reply to Maria: So if n is thousands ordered, then the order value per thousand is

   (30 - 0.375(n-50))

Multiply that by n to get the total revenue for the order.

   R(n)  =  n(30 - 0.375(n-50))  =  30n - 0.375n² + 18.75n

Simplify the expression on the right. Then find dR/dn by taking the derivative of the above. Set it to zero and solve for n. If you do everything right, you'll get the answer you've given.
Karl <Click to Send Email to Karl>
USA - Thursday, November 14, 2002 at 19:22:15 (EST)

Karl, Where am I approaching the problem incorrectly? I know how to take derivatives, and I can obviouslly set them equal to zero, and then plug them in to get the answer. My problem though is setting up the equations to work with. Do you have any advice for me? I'll be having a test on this stuff pretty soon (with in the next week or so), and it would be great if I could do well on it. Thanks again. (helpless)
Maria WH, FL USA - Tuesday, November 12, 2002 at 22:58:28 (EST)

Thanks for the help! Here's another one, if you have time, please help. (sorry, I'm only a highschool calculus student), therefore I'm not so good at this stuff. A company offers the following schedule of charges: $30 per thousand for orders of 50,000 or less, with the charge per thousand decreased by $.375 (37 1/2 cents) for each thousand above 50,000. Find the order size that makes the cpmpany's receipts a maximum. My book has an answer for this one, it is suppose to be 65,000.
Maria
wh , fl USA - Tuesday, November 12, 2002 at 22:51:16 (EST)

Reply to Maria : The problem tells you that the base of the box is square. So let the length of each side of that square be b. The only other dimension of the box is its height, h. The volume is given by

   V  =  b²h  =  32 in³

The area of material used is b² for the base, and bh for each of the four sides:

   A  =  b² + 4bh

From the volume equation you can solve for h in terms of b

   h  =  32 in² / b²

Substitute that into the area equation:

   A  =  b² + 4b(32/b²)  =  b² + 128/b

Now find dA/db and set it to zero:

   dA
       =  2b - 128/b²  =  0
   db

Now simply solve that for b, and then back-substitute to get h.
Karl <Click to Send Email to Karl>
USA - Tuesday, November 12, 2002 at 18:47:50 (EST)

I really need help with finding max and mins of word problems. I am so bad at those, and I could really use the extra help. Here's an example of the most current problem that I can not solve. A box with a square base and open top is to hold 32in^3. Find the dimensions that require the least amount of material. Neglect the thickness of the material and waste in the construction. I know you have to take the derivative and set it equal to zero, substitute and such. But I need help setting up the formula. Please help.
Maria WH, FL USA - Monday, November 11, 2002 at 22:35:23 (EST)

Ok I Tried the Chain rule and this is what I got. Could someone verify that this is correct. If it is not could you help me solve it and tell me what's wrong. D/Dx(a*cosx+(b^2 - a^2*sin^2x)^1/2) -a*sinx - (a^2 * 2*Sinx*Cosx)*(1/2*(b^2 - a^2 * Sin^2x)^-1/2) So how did I do. Thank you for you help Karl in your earlier reply. Todd
Todd Richardson Medina, Tn USA - Friday, November 08, 2002 at 19:17:21 (EST)

Reply to Zackery: You will find no algebraic or trigonometric trick with which to solve this one. There is no closed expression that gives you x in terms of y for

   y  =  tan(x) - x

In math-speak, we say that the function that takes y into x, in this case, is transcendental. But you still would like to solve it. There are numerical methods that, for a given y, will converge on the value of x that works. The fastest converging is Newton-Raphson. You have for a given y, the equation:

   tan(x) - x - y  =  0

where f(x) = tan(x) - x - y. Newton-Raphson says to pick a first guess, x₀, and do

   x_n+1  =  x_n - f(x_n)/f'(x_n)

In your case, f'(x) = sec²(x) - 1 (remember that y is to be considered a constant).

For y = 0.3 and x₀ = 1, I get

   x₁  =  0.893875189
   x₂  =  0.861230757
   x₃  =  0.858855626
   x₄  =  0.858844142

And x₄ is as accurate as my hand-calculator can get it. See how well it works for that y for yourself. Also try some different y's for yourself.
Karl <Click to Send Email to Karl>
USA - Thursday, November 07, 2002 at 21:41:59 (EST)

f(x) = tan(x)-x how do I solve for x in terms of y?
Zackery Rochester, NY USA - Thursday, November 07, 2002 at 18:46:26 (EST)

Reply to dr0dlr: I haven't a clue where you got that answer. This problem involves using the chain rule. You have

   f(t)  =  (1 + tan(t))^1/3

So the inner function is h(t) = 1 + tan(t). The outer function is g(h) = h^1/3. According to the chain rule you need to find g'(h(t))h'(t). The derivative of tan(t) is sec²(t). Use the power rule to find the derivative of h^1/3.

It is very important that you understand when and how to use the chain rule. The chain rule is central to things to come in your calculus course.
Karl <Click to Send Email to Karl>
USA - Wednesday, November 06, 2002 at 18:50:26 (EST)

Reply to todd richardson: I assume you want the derivative of that function. The derivative of sin(x) is cos(x), so that part is easy. For the rest of it, you have to apply the chain rule twice. The first time is to find the derivative of sin²(x). The second time to find the derivative of (b² - a² sin²(x))^(1/2), which is also a composite. Break each chain rule application up into an inner function and an outer function, and then apply the chain rule. For example, sin²(x) is a composite of the inner function, g(x) = sin(x) and the outer function, f(g) = g². So according to the chain rule, you need to find f'(g(x))g'(x). See if you can take it the rest of the way.
Karl <Click to Send Email to Karl>
USA - Wednesday, November 06, 2002 at 18:42:12 (EST)

Reply to S : If a series converges to a limit, L, then by definition, you can always go far enough into the series so that all partial sums taken to beyond that point are within e of L, no matter how close to zero you choose e. That means that all the partial sums to beyond that point must fall within e of each other. Can you see why that implies that the elements of the series must tend to zero themselves? Once you establish that it's clear that if s[n] converges, 1/s[n] cannot. Remember that s[n] is always the last term in a partial sum.
Karl <Click to Send Email to Karl>
USA - Wednesday, November 06, 2002 at 18:32:06 (EST)


Wondering if the answer to this problem ---

Differentiate f(t) = cubert(1 + tan(t))

The answer I concluded was  ---

sec^2(x^3)3x^2

Is that answer correct? If not please help me out.
Thanks


dr0dlr 



 USA - Wednesday, November 06, 2002 at 11:04:22 (EST)

I am sorry I made a mistake in the equation. Here is the correct version. d/dx[ a*sinx + (b^2 - a^2*sin^2x)^1/2 ] Thanks Todd
todd richardson medina, tn USA - Tuesday, November 05, 2002 at 22:44:42 (EST)

Hello Everyone It has been along time since I have had to do differential equations. I have come across this equation that I am trying to solve and am having problems. I was wondering if someone can help me solve it. I'd like to be taught how it is solved not just the answer. d/dx[a*cosx + (b^2 + a^2*sin^2x)^1/2] a and b are constants. Any help would be greatly appreciated. My email is rarichar411@aol.com Thank you Todd
todd richardson Medina, tn USA - Tuesday, November 05, 2002 at 22:16:34 (EST)

Given: The sum of a series, A sub n from n=1 to infinity is convergent and A sub n is not equal to 1. Prove that the sum of a series, one divided by A sub n is divergent. Does anyone know how to prove this?
S IF, ID USA - Tuesday, November 05, 2002 at 19:00:13 (EST)

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