Multivariable calculus question

[kwal0203]

Homework Statement

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.

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!

 

[vela]

Homework Statement

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.

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)

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

(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!

 

[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})

 

[vela]

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

 

[kwal0203]

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

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)?

 

[vela]

For iia?

(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!

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?

 

[kwal0203]

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?

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

 

[vela]

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.