At the point (2,-1,2) on the surface z = x(y^2), find the direction vector for the direction of greatest decrease of z.(adsbygoogle = window.adsbygoogle || []).push({});

A) i - 2j

B) i - 4j

C) (i - 4j)/sqrt(17)

D) -i + 4j

E) i + j

Do I need a function f(x,y,z)? If so then f(x,y,z) = z = xy^2.

Then the gradient vector would be <Fx, Fy, Fz>.

This would be give me. < -y^2, -2xy, 1>

At the point (2,-1,2) we get < -1, 4, 1>

But how do i get the direction of the vector of greatest decrease?

From the choices though it seems they only find the gradient vector of z = F(x,y) = xy^2

In which case the gradient vector is < fx, fy>

< y^2 , 2xy> .

At the point (2, -1 ,2) we get < 1, -4>

Again i am not sure how to find the vector of greatest decrease.

It says in my book the direction of minimum increase is - norm( grad f(x,y,z)). But i am not getting any of these answers can some one help?

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# Homework Help: Calculus problem with directional vectors

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