## Finding the gradient and maximum gradient at a point P for the surface z=2x^2+3y^2

1. The problem statement, all variables and given/known data
For the surface z=2x^2+3y^2, find
(i) the gradient at the point P (2,1,11) in the direction making an angle a with the x-axis;
(ii) the maximum gradient at P and the value of a for which it occurs.

2. Relevant equations
ma=$\partial$z/$\partial$x(cos(a))+$\partial$z/$\partial$y(sin(a))

If dma/da = 0 and d^2ma/da^2<0 the ma, is a maximum for that value of a

3. The attempt at a solution\partial

(i) Firstly I calculated:
$\partial$z/$\partial$x = 4x
$\partial$z/$\partial$y = 6y

Therefore applying equation: ma=$\partial$z/$\partial$x(cos(a))+$\partial$z/$\partial$y(sin(a)) at P(2,1,11)

We get:

ma=8cos(a)+6sin(a)
(I am quite sure this is correct, I don't have answers so if I am doing something wrong can someone please inform me.

(ii) This is where I am stuck, I can't quite get the correct working out, can anyone please help out, below is the calculations I attempted:

If dma/da = 0 and d^2ma/da^2<0 the ma, is a maximum for that value of a.

dma/da = -8sin(a)+6cos(a)

dma/da=0 Therefore: -8sin(a) + 6cos(a) = 0 Therefore: 8sin(a)=6cos(a)

This is what I was up to, got stuck here, not sure if what I done is correct, if it is, how do I proceed from here?

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I have now calculated part (ii), I realised I have a problem in part (i), I need to find a to get the gradient ma, can anybody help me out with this?

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Recognitions:
Homework Help
 Quote by savva Therefore applying equation: ma=$\partial$z/$\partial$x(cos(a))+$\partial$z/$\partial$y(sin(a)) at P(2,1,11) We get: ma=8cos(a)+6sin(a) (I am quite sure this is correct, I don't have answers so if I am doing something wrong can someone please inform me.
It is correct for the derivative at the given point along the direction defined by the angle α.

 Quote by savva (ii) This is where I am stuck, I can't quite get the correct working out, can anyone please help out, below is the calculations I attempted: If dma/da = 0 and d^2ma/da^2<0 the ma, is a maximum for that value of a. dma/da = -8sin(a)+6cos(a) dma/da=0 Therefore: -8sin(a) + 6cos(a) = 0 Therefore: 8sin(a)=6cos(a) This is what I was up to, got stuck here, not sure if what I done is correct, if it is, how do I proceed from here?

Isolate α: sin(α)/cos(α)=6/8 =tan(α)=0.75 α=?
Getting α, substitute it back to mα.

ehild