Multivariable max rate of change

In summary: Good luck on your midterm!In summary, the equation z(e^(xy)) + z^5 + y = 4 implicitly defines z as a function z = f(x,y) near (0,2,1). To find df/dx and df/dy, use the product rule and implicit differentiation. For part (b), find the maximum value of the gradient by substituting the given point (0,2,1) into the gradient of f. To implicitly differentiate z with respect to x, differentiate every term with respect to x, noting that y does not depend on x and z does. Good luck on your midterm!
  • #1
allenh98
3
0

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)
 
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  • #2
allenh98 said:

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

[tex]ze^{xy} + z^5 + y = 4[/tex]

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 zx. Similarly for zy.

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?
 
  • #3
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.
 
  • #4
allenh98 said:
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

[tex]
ze^{xy} + z^5 + y = 4
[/tex]
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:

[tex]z_xe^{xy} + zye^{xy} + 5z^4z_x + 0 = 0[/tex]

and solve for zx.
 

What is multivariable max rate of change?

Multivariable max rate of change refers to the maximum rate at which a function changes with respect to multiple variables. It is a measure of the steepest slope or the fastest rate of change in a multivariable function.

Why is multivariable max rate of change important?

Multivariable max rate of change is important in many fields of science, such as physics, economics, and engineering. It helps us understand how multiple variables affect a system and allows us to optimize processes and solve real-world problems.

How is multivariable max rate of change calculated?

The multivariable max rate of change is calculated by finding the partial derivatives of the function with respect to each variable and then setting them equal to zero. The resulting values are then used to determine the maximum rate of change.

What is the difference between multivariable max rate of change and single variable max rate of change?

The main difference between multivariable max rate of change and single variable max rate of change is that the former considers the impact of multiple variables on a function, while the latter only considers one variable. Multivariable max rate of change is also more complex to calculate and has a wider range of applications.

What are some real-world applications of multivariable max rate of change?

Multivariable max rate of change is used in various fields, such as determining the optimal production levels in manufacturing, finding the best routes for transportation, and predicting changes in stock prices. It is also used in physics to analyze the motion of objects in multiple dimensions.

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