Mark's Calc III - Individual question feed

Directional Derivative equal to 10

SpaceManSpiff — Thu, 17 Jul 2014 15:43:34 -0500

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?

Exam 2 Review Sheet

Anonymous — Wed, 16 Jul 2014 07:09:35 -0500

Number 4, part c, of the exam 2 review sheet asks: Let $f(x,y)=xy^3$ (c) From the point $(2,1)$, is there any direction $u$ so that $D_uf(2,1)=10$? So far I have calculated the gradient to be $$\bigtriangledown f(2,1)=<8,6>$$ The idea I had was to create a dot product of $\bigtriangledown f(2,1)$ and a general $u$ vector, and set it equal to 10. The trouble I'm having is figuring out how to solve the two unknowns because I only have one equation.

Gradient of f(x,y)

SpaceManSpiff — Wed, 16 Jul 2014 20:36:25 -0500

So I feel like I should know this but... In three dimensional space the del operator is defined as $\nabla =\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \rangle$. So when dealing with a function like the one in problem 4 of the review sheet; $f(x,y)=xy^{3}$. Do we treat del as being $\nabla=\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \rangle$ and find a gradient with two components or do we remake the function into a level surface such as $xy^{3}-z=k $ where $k \in \mathbb{Z} $ and find a gradient with three components? If this is obvious I'm incredibly sorry but it really is bugging me.

I'm lost on using the gradient for tangent planes.

SpaceManSpiff — Wed, 09 Jul 2014 23:43:53 -0500

Can someone list a step by step procedure for utilizing the gradient of some $f(x)$ to find the tangent plane to a surface? I'm confused about the distinction between surfaces and level surfaces also. *Comment*: It would be *much* easier to provide a reasonable answer if you provided a specific problem that you're working on with a specific function $f$.