# Multivariable calculus question

1. Mar 2, 2013

### kwal0203

1. The problem statement, all variables and given/known data

A function $f(x,y)=\mathbb{R}^{2}\rightarrow \mathbb{R}$ is defined by:

$f(x,y)=5y^{2}-x^{2}$

(i) Find its gradient vector and evaluate it at the point $(x,y)^{T}=(1,1)^{T}$. Find the rate of change of the function in the direction $(2,1)^{T}$ at the point $(1,1)^{T}$

(ii) In what direction is the rate of change of the function at the point $(x,y)^{T}=(2,0)^{T}$ equal to $-2$?

Is there a direction for which the rate of change at the point $(x,y)^{T}=(2,0)^{T}$ is equal to $-5$? Find the greatest rate of decrease of the function at this point.

3. The attempt at a solution

(i)

$f(x,y)=5y^{2}-x^{2}$

$f_{x}(x,y)=-2x=-2$ at (1,1)

$f_{y}(x,y)=10y=10$ at (1,1)

The unit vector in the direction of $(2,1)^{T}$ is $(2,1)^{T}=\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}$

To find the rate of change in the direction of the unit vector above I:

$<\bigtriangledown f,u>=f_{x}u_{1}+f_{y}u_{2}=-2\cdot \frac{2}{\sqrt{5}}+10\cdot \frac{1}{\sqrt{5}}=\frac{6}{\sqrt{5}}$

(ii)

at $(2,0)^{T}$ the gradient vector is:

$\bigtriangledown f=(-4,0)$

so knowing $(\bigtriangledown f,u)=-2$ gives

$-2=-4\cdot u_{1}+0\cdot u_{2}$

$-2=-4\cdot u_{1}$

$u_{1}=\frac{1}{2}$

so the direction in which the rate of change of the function is $u=(\frac{1}{2},0)$

(iia) for the part where it asks about the rate of change being $-5$ I used the same method and got $u=(\frac{5}{4},0)$.

Is this all correct?

but now I don't know how to 'Find the greatest rate of decrease of the function at this point.'

Any help appreciated thanks!

Last edited: Mar 2, 2013
2. Mar 2, 2013

### vela

Staff Emeritus
Actually, you don't know that $u_2=0$. You want to use the fact that $\vec{u}$ is a unit vector to determine $u_2$.

3. Mar 2, 2013

### kwal0203

Oh I see, so:

$\parallel u \parallel=1=\sqrt{\frac{1}{2}\cdot \frac{1}{2}+u_{2}^{2}}$

$\frac{1}{4}+u_{2}^{2}=1^{2}$

$u_{2}^{2}=1-\frac{1}{4}$

$u_{2}=\sqrt{\frac{3}{4}}$

$u_{2}=\pm \frac{1}{2}\sqrt{3}$

$u=(\frac{1}{2},\pm \frac{1}{2}\sqrt{3})$

4. Mar 2, 2013

### vela

Staff Emeritus
Yup, so can you see now that the first answer to iia is wrong?

5. Mar 2, 2013

### kwal0203

I think so:

$u=(\frac{1}{2},\pm \frac{1}{2}\sqrt{3})$

$\bigtriangledown f=(-4,0)$

so,

$(\bigtriangledown f,u)=-4\cdot \frac{1}{2}+0$

$u=(-2,0)$???

6. Mar 3, 2013

### vela

Staff Emeritus
For iia?

The gradient evaluated at the point is (-4,0), and now you're dotting a unit vector into it. When is the dot product the greatest? What is the biggest value $(-4,0)\cdot\vec{u}$ can take on when $\vec{u}$ is a unit vector?

7. Mar 3, 2013

### kwal0203

I'm not sure. When is the dot product the greatest?

You mean when $\left \| \bigtriangledown f \right \|=\sqrt{-4^{2}+0}$

$\left \| \bigtriangledown f \right \|=4$?

then,

$-\left \| \bigtriangledown f \right \|=-4$?

so,

$-4$ is the minimum and there is no direction that can make the rate of change $-5$

Last edited: Mar 3, 2013
8. Mar 4, 2013

### vela

Staff Emeritus
Yeah, pretty much.

The dot product of two vectors $\vec{A}$ and $\vec{B}$ is given by $\vec{A}\cdot\vec{B} = \|\vec{A}\| \|\vec{B}\| \cos \theta$, where $\theta$ is the angle between them. Clearly, the product must be between $-\|\vec{A}\| \|\vec{B}\|$ and $\|\vec{A}\| \|\vec{B}\|$. In this problem, this means that the directional derivative must be between -4 and +4.