Multivariable calculus. The chain rule.

In summary, the partial derivatives for z are:\partialz/\partialu = (2xuy sin(u) tan(v))(x\partialx/\partialu + y\partialy/\partialu + cos(u)tan(v))\partialz/\partialv = (2xuy sin(u) tan(v))(x\partialx/\partialv + y\partialy/\partialv + sin(u)sec^2(v)) where x = x(u,v), y = y(u,v), and x = 2.112, y = 4.797, \partialx/\partialu = -3.491, \partialx/\partialv = -2.230 , \partialy/\partialu
  • #1
letalea
8
0

Homework Statement



Let x=x^2ysin(u)tan(v), where x(u,v) and y(u,v) are smooth functions that, when evaluated at u=1 and v=-3 satisfy

x=2.112, y=4.797, [tex]\partial[/tex]x/[tex]\partial[/tex]u = -3.491, [tex]\partial[/tex]x/[tex]\partial[/tex]v = -2.230 , [tex]\partial[/tex]y/[tex]\partial[/tex]u = 1.787 , [tex]\partial[/tex]y/[tex]\partial[/tex]v = 1.554.

Then the value of [tex]\partial[/tex]z/[tex]\partial[/tex]u + ln ( [tex]\partial[/tex]z/[tex]\partial[/tex]v), at u =1, v= -3 is one of the following:

9.004
-1.225
-3.257
3.896
6.469
-2.368
-9.311
-3.658


The Attempt at a Solution



[tex]\partial[/tex]z/[tex]\partial[/tex]u= (2(2.112)(4.797)sin(1)tan(-3))(-3.491) + (2.112^2 sin(1)tan(2.112))(1.787) = 0.057407948

[tex]\partial[/tex]z/[tex]\partial[/tex]v=(2(2.112)(4.797)sin(1)tan(-3))(-2.230) + (2.112^2 sin(1)tan(-3))(1.554) = 0.034988483

0.057407948 + ln(0.034988483) = -3.295


The answer, which you can tell, is not even a solution on the list of possible answers. Any insight to where I went wrong would be appreciated! Thanks in advance!
 
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  • #2
i think that you write some formulas wrong (you have mistake in writing) .
 
  • #3
letalea said:

Homework Statement



Let x=x^2ysin(u)tan(v)
This would make more sense as z = x2y sin(u) tan(v).
letalea said:
, where x(u,v) and y(u,v) are smooth functions that, when evaluated at u=1 and v=-3 satisfy

x=2.112, y=4.797, [tex]\partial[/tex]x/[tex]\partial[/tex]u = -3.491, [tex]\partial[/tex]x/[tex]\partial[/tex]v = -2.230 , [tex]\partial[/tex]y/[tex]\partial[/tex]u = 1.787 , [tex]\partial[/tex]y/[tex]\partial[/tex]v = 1.554.

Then the value of [tex]\partial[/tex]z/[tex]\partial[/tex]u + ln ( [tex]\partial[/tex]z/[tex]\partial[/tex]v), at u =1, v= -3 is one of the following:

9.004
-1.225
-3.257
3.896
6.469
-2.368
-9.311
-3.658


The Attempt at a Solution



[tex]\partial[/tex]z/[tex]\partial[/tex]u= (2(2.112)(4.797)sin(1)tan(-3))(-3.491) + (2.112^2 sin(1)tan(2.112))(1.787) = 0.057407948

[tex]\partial[/tex]z/[tex]\partial[/tex]v=(2(2.112)(4.797)sin(1)tan(-3))(-2.230) + (2.112^2 sin(1)tan(-3))(1.554) = 0.034988483

0.057407948 + ln(0.034988483) = -3.295


The answer, which you can tell, is not even a solution on the list of possible answers. Any insight to where I went wrong would be appreciated! Thanks in advance!
 
  • #4
Yes sorry, it was a posting mistake, the question was z=x^2ysin(u)tan(v). Can you see any errors in how I solved the partial derivatives, as I have tried multiple times and can not seem to find the correct solution.
 
  • #5
So you have z = x2(u, v) y(u, v) sin(u) tan(v).

Before putting in all the numbers, what do you get for zu and zv?

zu is just another way of writing [tex]\frac{\partial z}{\partial u}[/tex]
 
  • #6
Okay I hope this is what you were wanting me to answer...

[tex]\partial[/tex]z/[tex]\partial[/tex]u = ([tex]\partial[/tex]z/[tex]\partial[/tex]x) ([tex]\partial[/tex]x/[tex]\partial[/tex]u) + ([tex]\partial[/tex]z/[tex]\partial[/tex]y)([tex]\partial[/tex]y/[tex]\partial[/tex]u)

=(2xysin(u)tan(v))([tex]\partial[/tex]x/[tex]\partial[/tex]u) +(x^2sin(u)tan(v))([tex]\partial[/tex]y/[tex]\partial[/tex]u)

[tex]\partial[/tex]z/[tex]\partial[/tex]v= ([tex]\partial[/tex]z/[tex]\partial[/tex]x) ([tex]\partial[/tex]x/[tex]\partial[/tex]v) + ([tex]\partial[/tex]z/[tex]\partial[/tex]y)([tex]\partial[/tex]y/[tex]\partial[/tex]v)

=(2xysin(u)tan(v))([tex]\partial[/tex]x/[tex]\partial[/tex]v) + (x^2sin(u)tan(v))([tex]\partial[/tex]y/[tex]\partial[/tex]v)
 
  • #7
That looks OK. It will be easier to evaluate all the functions and partials using subscript notation.

So zu = 2xysin(u)tan(v) xu + x2sin(u)tan(v) yu

and zv = 2xysin(u)tan(v) xv + x2sin(u)tan(v) yv


You have x(1, -3) = 2.112, y(1, -3) = 4.7997
xu(1, -3) = -3.491
xv(1, -3) = -2.230
yu(1, -3) = 1.787
yv(1, -3) = 1.554

You need to evaluate zu(1, -3) + ln(zv(1, -3))

Plug the function/partial derivative values in. Make sure your calculator is in radian mode when you evaluate sin(1) and tan(-3).
 
Last edited:
  • #8
Thank you I will give that a try!
 
  • #9
I am working on this question as well. Though, upon evaluating ln(Zv) I came out with a negative number, and ln of a negative is not valid. What did I do wrong?
ln(2(2.112)(4.797)sin(1)tan(-3)(-2.230) + ((2.112^2)sin(1)tan(-3)(1.554))) = Zv

I'm certain the partail derivative is correct, I've gone over it and tried to manipulate different ways, but still not the right answer.
 
  • #10
I didn't check letalea's work closely back in post #7, and the partial derivatives shown there are missing a term in each of them.

Assuming z = x2y * sin(u) * tan(v), where x = x(u, v) and y = y(u, v) as described earlier, then

zu = 2x * xu * y * sin(u) * tan(v) + x2 *yu * sin(u) * tan(v) + x2 * y * cos(u) * tan(v)

In letalea's work, that last term is missing.

Also,
zv = 2x * xv * y * sin(u) * tan(v) + x2 *yv * sin(u) * tan(v) + x2 * y * sin(u) * sec2(v)

The last term is missing in this partial, as well.

For a sanity check, evaluate zv at (1, -3) to see if it comes up positive this time (so you can take its log).
 
  • #11
Thank you, that was very helpful! I was wondering if I needed to evaluate the trig function in relation to u! :)
 

Related to Multivariable calculus. The chain rule.

1. What is multivariable calculus?

Multivariable calculus is a branch of mathematics that deals with the study of functions of multiple variables. It involves the analysis of curves, surfaces, and other objects in higher dimensions.

2. What is the chain rule in multivariable calculus?

The chain rule in multivariable calculus is a rule that allows us to find the derivative of a composite function. It states that the derivative of a composite function is equal to the product of the derivative of the outer function and the derivative of the inner function.

3. How is the chain rule applied in multivariable calculus?

To apply the chain rule in multivariable calculus, we first need to identify the outer function and the inner function in the composite function. Then, we take the derivative of the outer function, leaving the inner function unchanged. Finally, we multiply this result by the derivative of the inner function.

4. Why is the chain rule important in multivariable calculus?

The chain rule is important in multivariable calculus because it allows us to find the derivatives of complex functions. Many physical and mathematical phenomena can be described by composite functions, and the chain rule is essential in understanding and analyzing these phenomena.

5. What are some common applications of the chain rule in multivariable calculus?

The chain rule has various applications in fields such as physics, engineering, economics, and computer science. It is used to find the rate of change and optimization of multivariable functions, as well as in the study of motion, vectors, and differential equations.

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