Multivariable calculus. The chain rule.

[letalea]

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!

 

[m.medhat]

i think that you write some formulas wrong (you have mistake in writing) .

 

[Mark44]

Homework Statement

Let x=x^2ysin(u)tan(v)

This would make more sense as z = x²y sin(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!

 

[letalea]

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.

 

[Mark44]

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 z_u and z_v?

z_u is just another way of writing [tex]\frac{\partial z}{\partial u}[/tex]

 

[letalea]

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)

 

[Mark44]

That looks OK. It will be easier to evaluate all the functions and partials using subscript notation.

So z_u = 2xysin(u)tan(v) x_u + x²sin(u)tan(v) y_u

and z_v = 2xysin(u)tan(v) x_v + x²sin(u)tan(v) y_v


You have x(1, -3) = 2.112, y(1, -3) = 4.7997
x_u(1, -3) = -3.491
x_v(1, -3) = -2.230
y_u(1, -3) = 1.787
y_v(1, -3) = 1.554

You need to evaluate z_u(1, -3) + ln(z_v(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).

 

[letalea]

Thank you I will give that a try!

 

[Dee247]

I am working on this question as well. Though, upon evaluating ln(Z<sub>v) 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.</sub>

 

[Mark44]

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

z_u = 2x * x_u * y * sin(u) * tan(v) + x² *y_u * sin(u) * tan(v) + x² * y * cos(u) * tan(v)

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

Also,
z_v = 2x * x_v * y * sin(u) * tan(v) + x² *y_v * 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 z_v at (1, -3) to see if it comes up positive this time (so you can take its log).

 

[Dee247]

Thank you, that was very helpful! I was wondering if I needed to evaluate the trig function in relation to u! :)