# Math Help - Finding directional derivative, tangent and path of maximal increase

1. ## Finding directional derivative, tangent and path of maximal increase

Hi, I have the following question:

Consider the function f(x,y,z)=x^3+y^4-z^3

1) Find the points on the surface f(x,y,z)=1 that have horizontal tangent plane (tangent plane parallel to the xy plane).

This is the question I am unsure of.
However I have:
grad f=(3x^2)i+(4y^3)j-(3z^2)k

2) Compute the directional rerivative at point (0,1,1) in the direction of u=i+j+k

I found this answer to be 1/sqrt(3)

3) Suppose G(x,y)=f(x,y,0)=x^3+y^4 describes the profit of a company as a measure of two variables. What should be your best strategy to modify y as a function of x in order to maximise the profit if you know that initially x=1, y=1

I got the path of maximal increase to be y=(y^4)/(3x^2) + (2/3)
However I am unsure what to say about modifying y.

Cheers.

2. Originally Posted by BlueEagle
Hi, I have the following question:

Consider the function f(x,y,z)=x^3+y^4-z^3

1) Find the points on the surface f(x,y,z)=1 that have horizontal tangent plane (tangent plane parallel to the xy plane).

This is the question I am unsure of.
However I have:
grad f=(3x^2)i+(4y^3)j-(3z^2)k

2) Compute the directional rerivative at point (0,1,1) in the direction of u=i+j+k

I found this answer to be 1/sqrt(3)
The directional derivative at a point $p = (0,1,1)$ in the direction $u = (1,1,1)$ is just $\nabla f(p) \cdot u = (0,4,-3)^T(1,1,1) = 1.$

3. Originally Posted by JakeD
The directional derivative at a point $p = (0,1,1)$ in the direction $u = (1,1,1)$ is just $\nabla f(p) \cdot u = (0,4,-3)^T(1,1,1) = 1.$
Aren't you supposed to use the unit vector of u?

4. Originally Posted by JakeD
The directional derivative at a point $p = (0,1,1)$ in the direction $u = (1,1,1)$ is just $\nabla f(p) \cdot u = (0,4,-3)^T(1,1,1) = 1.$
Originally Posted by BlueEagle
Aren't you supposed to use the unit vector of u?
OK, I agree. I gave the definition of the directional derivative along u instead of the directional derivative in the direction of u.

5. Originally Posted by BlueEagle
Hi, I have the following question:

Consider the function f(x,y,z)=x^3+y^4-z^3

1) Find the points on the surface f(x,y,z)=1 that have horizontal tangent plane (tangent plane parallel to the xy plane).

This is the question I am unsure of.
However I have:
grad f=(3x^2)i+(4y^3)j-(3z^2)k
A tangent plane parallel to the xy plane has unit normal (0,0,1). The gradient is normal to the tangent plane. So grad f must be a multiple of (0,0,1), implying x = y = 0, f(0,0,z) = 1, -z^3 = 1 and thus z = -1. Hence (0,0,-1) is the desired point.