Directional Derivative equal to 10

asked 2014-07-17 15:43:34 -0600

SpaceManSpiff
703 ●4 ●22 ●40

updated 2014-07-17 16:24:10 -0600

So with # 4 on the review sheet. $$\nabla f=\langle y^{3},3xy^{2} \rangle$$ So at $(1,2)$ $\nabla f=\langle1,6 \rangle.$

So to determine if $D_{u}f$ is equal to $10$ we solve for $a$ and $b$ in the equation $$\langle1,6 \rangle \cdot \langle a,b \rangle=10$$

Here's where I feel confused. You end up with $a+6b=10.$ Now $a=4, b=1$ is an obvious solution to this equation. However, does $\textbf{u}$ have to be a unit vector to correctly answer this question?

This leads to two other questions.

If it does need to be unit vector do we find it by normalizing the vector $\langle4,1 \rangle$? If we do normalize this vector the components of the unit vector we construct will not solve the equation $a+6b=10.$ But I feel that since it will still have the same direction this may not matter.

As an alternative method we could solve the system of equations below? $$ a+6b=10$$ $$a^{2}+b^{2}=1$$

I haven't bothered trying to solve that system yet because I'm not sure it's necessary. Thoughts?

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answered 2014-07-17 16:34:19 -0600

Dylan
650 ●2 ●16 ●25

updated 2014-07-17 16:56:07 -0600

I think this is the problem with using that vector: The question is asking if there is a direction where the magnitude of the rate of change is 10, and the magnitude of the rate of change is given by the magnitude of the gradient vector. If we dot the grad vector with a non-unit vector, the dot product isn't equal to the magnitude of the rate of change, so sadly non-unit vectors don't work. In general, the directional derivative needs a unit vector for this reason.

If we normalize the vector, then it no longer makes the dot product equal to 10, so unfortunately we can't do that either. I think the only way to find it methodically is the solve the system of equations.

Edit in response to Christina: Ah, so if we had been asked to "find the direction where the rate of change is 10" we'd have to solve the system, but because we were only asked if it ever had that value, we can just see if it ever goes that high. I completely forgot that day in class.

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answered 2014-07-17 16:23:23 -0600

Justin
1009 ●2 ●20 ●38

updated 2014-07-18 13:36:23 -0600

Take a look at my answer to a previous question.

Update/Comment: Thanks Christina! I think you explain things wonderfully!

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answered 2014-07-17 16:25:32 -0600

Christina
807 ●18 ●35 https://sites.google.com/...

I think Justin and Tiffany did a nice job addressing this one previously. You don't really need to set up systems to solve. If we know the gradient vector points toward the direction with the greatest change, that is all you need.

If $$\nabla f=\langle y^{3},3xy^{2} \rangle$$

Then $$\nabla f(2,1)=\langle 1,6 \rangle$$

And we know this is the direction of the greatest change. So, what is the change? The magnitude of the gradient. $$|| \ \nabla f \ || = \sqrt{1^2 + 6^2} = \sqrt{37} $$

Now all you have to do is compare this to 10 and see that it will never be that high. I hope this helps, I am not as elegant as Justin at explaining....

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Directional Derivative equal to 10

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