Rate of change (partial differentiation)

• March 28th 2014, 05:36 AM
kaemper
Rate of change (partial differentiation)
Greetings,

Attachment 30539I am almost done with this exercise, but I don't get a sqrt(5) in the denominator in my final answer, thus the correct answer is: 5/sqrt(5). How come I don't get a correct answer? Is the formula used incorrect?
• March 28th 2014, 07:19 AM
hollywood
Re: Rate of change (partial differentiation)
When you calculate a directional derivative, you need to do the inner product with a unit vector. The unit vector in the direction $\vec{i}+2\vec{j}$ is $\frac{1}{\sqrt{5}}\vec{i}+\frac{2}{\sqrt{5}}\vec{j }$. So when you take the dot product, you should get $\frac{4}{\sqrt{5}}$. I assume your answer 5/sqrt(5) is a typo.

- Hollywood
• March 28th 2014, 10:43 AM
kaemper
Re: Rate of change (partial differentiation)
Yes, it is certainly a mistake.
• March 29th 2014, 07:19 PM
HallsofIvy
Re: Rate of change (partial differentiation)
You also have the gradient itself wrong. The gradient of $x^2y$ at (-1, -1) is NOT $2\vec{i}+ \vec{j}$.
• March 30th 2014, 09:41 PM
hollywood
Re: Rate of change (partial differentiation)
It's not? I calculated $\nabla(x^2y) = \vec{i}\frac{\partial}{\partial{x}}(x^2y) + \vec{j}\frac{\partial}{\partial{y}}(x^2y) = 2xy\vec{i}+x^2\vec{j}$, and substituting $x=y=-1$ gives $2\vec{i}+\vec{j}$.

What am I missing?

- Hollywood
• August 13th 2014, 02:24 PM
kaemper
Re: Rate of change (partial differentiation)
Quote:

Originally Posted by HallsofIvy
The gradient of $x^2y$ at (-1, -1) is NOT $2\vec{i}+ \vec{j}$.

Why is it so, HallsofIvy. The two of us cannot understand that.

Respectfully,
Kaemper
• August 13th 2014, 02:45 PM
Cartesius24
Re: Rate of change (partial differentiation)
I agree with previous (we say about unnormed vector)
• August 13th 2014, 03:16 PM
kaemper
Re: Rate of change (partial differentiation)
Quote:

Originally Posted by Cartesius24
I agree with previous (we say about unnormed vector)

Do you agree with HallsofIvy?
• August 13th 2014, 03:25 PM
Cartesius24
Re: Rate of change (partial differentiation)
With You
• August 13th 2014, 03:32 PM
HallsofIvy
Re: Rate of change (partial differentiation)
Quote:

Originally Posted by hollywood
It's not? I calculated $\nabla(x^2y) = \vec{i}\frac{\partial}{\partial{x}}(x^2y) + \vec{j}\frac{\partial}{\partial{y}}(x^2y) = 2xy\vec{i}+x^2\vec{j}$, and substituting $x=y=-1$ gives $2\vec{i}+\vec{j}$.

What am I missing?

- Hollywood

Okay, I was wrong. For some reason I didn't realize that we were multiplying two negatives!