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

1. Jan 17, 2012

### savva

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?

Last edited: Jan 17, 2012
2. Jan 17, 2012

### savva

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

$\int$

3. Jan 17, 2012

### ehild

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

It is correct for the derivative at the given point along the direction defined by the angle α.

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

ehild