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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.

It 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 it 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?

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.

It 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 it 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?

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.

It 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 it 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?

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 it 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?