# Thread: Rate of change (partial differentiation)

1. ## Rate of change (partial differentiation)

Greetings,

I 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?

2. ## 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 $\displaystyle \vec{i}+2\vec{j}$ is $\displaystyle \frac{1}{\sqrt{5}}\vec{i}+\frac{2}{\sqrt{5}}\vec{j }$. So when you take the dot product, you should get $\displaystyle \frac{4}{\sqrt{5}}$. I assume your answer 5/sqrt(5) is a typo.

- Hollywood

3. ## Re: Rate of change (partial differentiation)

Yes, it is certainly a mistake.

4. ## Re: Rate of change (partial differentiation)

You also have the gradient itself wrong. The gradient of $\displaystyle x^2y$ at (-1, -1) is NOT $\displaystyle 2\vec{i}+ \vec{j}$.

5. ## Re: Rate of change (partial differentiation)

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

What am I missing?

- Hollywood

6. ## Re: Rate of change (partial differentiation)

Originally Posted by HallsofIvy
The gradient of $\displaystyle x^2y$ at (-1, -1) is NOT $\displaystyle 2\vec{i}+ \vec{j}$.
Why is it so, HallsofIvy. The two of us cannot understand that.

Respectfully,
Kaemper

7. ## Re: Rate of change (partial differentiation)

I agree with previous (we say about unnormed vector)

8. ## Re: Rate of change (partial differentiation)

Originally Posted by Cartesius24
I agree with previous (we say about unnormed vector)
Do you agree with HallsofIvy?

With You

10. ## Re: Rate of change (partial differentiation)

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

What am I missing?

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