Multivariable max rate of change

[allenh98]

Homework Statement

The equation z(e^(xy)) + z^5 + y = 4 implicitly defines z as a function z = f(x,y) near (0,2,1)
(a) find df/dx and df/duy where x = 0, y = 0, and z = 1
(b) find the maximum rate of change of f at the point (0,2)

Homework Equations

sorry, my first post here not sure what i need to write in this section

The Attempt at a Solution

sol'n (a):
df/dx = yz(e^(xy))
df/dy = xz(e^(xy)) + 1
then i sub in the co-ords given to find
df/dx = 2
df/dy = 1

part (b) is where I am confused, since they only give us f at point (0,2), is it implied that we plug in f(0,2) and get z = 1 and solve the gradient?
here is what i did:
grad(f) = [yze^(xy)]i + [xe^(xy) + 1]j + [e^(xy) + 5z^4]k
substitute (0,2,1)
2i + j + 6k
= sqrt(41)

 

[LCKurtz]

Homework Statement

The equation z(e^(xy)) + z^5 + y = 4 implicitly defines z as a function z = f(x,y) near (0,2,1)
(a) find df/dx and df/duy where x = 0, y = 0, and z = 1

You mean (0,2,1). (0,0,1) isn't on the surface.

(b) find the maximum rate of change of f at the point (0,2)

Homework Equations

sorry, my first post here not sure what i need to write in this section

The Attempt at a Solution

sol'n (a):
df/dx = yz(e^(xy))
df/dy = xz(e^(xy)) + 1

These partials aren't calculated correctly. Remember that z is implicitly a function of x and y. So if you differentiate

ze^{xy} + z^5 + y = 4

with respect to x, you must use the product rule on the first term and you must differentiate the z⁵ implicitly with respect to x. Then solve for z_x. Similarly for z_y.

part (b) is where I am confused, since they only give us f at point (0,2), is it implied that we plug in f(0,2) and get z = 1 and solve the gradient?
here is what i did:
grad(f) = [yze^(xy)]i + [xe^(xy) + 1]j + [e^(xy) + 5z^4]k
substitute (0,2,1)
2i + j + 6k
= sqrt(41)

Almost OK on part b. But 2i + j + 6k is not equal to sqrt(41). One is a vector and the other is a scalar. What did you mean to write there?

 

[allenh98]

Thank you for your help! I have my midterm in 3 hours lol. I meant to write the maximum value of the gradient, not direction. We haven't learned implicit partial differentiation, so I assume it will not be asked on my test. But just in case it is, how do I implicitly differentiate z wrt x?

And yes, you are correct it isn't (0,0,1) it was a typo sorry.

 

[LCKurtz]

Thank you for your help! I have my midterm in 3 hours lol. I meant to write the maximum value of the gradient, not direction. We haven't learned implicit partial differentiation, so I assume it will not be asked on my test. But just in case it is, how do I implicitly differentiate z wrt x?

To differentiate

<br /> ze^{xy} + z^5 + y = 4<br />
with respect to x you would differentiate every term with respect to x, noting that y doesn't depend on x and z does, and write:

z_xe^{xy} + zye^{xy} + 5z^4z_x + 0 = 0

and solve for z_x.