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

Find the direction

Fri, 11 Jul 2014 07:23:36 -0500

I am stumped on how to begin question #16 in section 14.5: Find the directions in which the directional derivative of $f(x,y)=x^2+sin(xy)$ at the point $(1,0)$ has the value 1. I'm sure $<0,1>$ would work but I don't see how to find all other directions.

Answer by Dylan for Find the direction

Fri, 11 Jul 2014 08:48:24 -0500

If you go through the process of computing the directional derivative, you should have the gradient of $f$ dotted with $\hat u$: $$D_{\hat u} = \nabla f \cdot \hat u$$ Here, $\nabla f = \langle 2, 1 \rangle$, and let's say that $\hat u = \langle u_1, u_2 \rangle$. Then the directional derivative is $$\langle 2,1 \rangle \cdot \langle u_1, u_2 \rangle$$ $$= 2u_1 + u_2 $$ We want the direction with the derivative equal to one, so we should set this whole thing equal to one: $$2u_1 + u_2 = 1$$ However, the vector also has to be of unit-length, so we know that $u_1^2 + u_2^2 =1$. Now we have two equations and two unknowns; the solutions to this system should give you the components of $\hat u$ that work. I found another solution on Mathematica, but I'm not sure how you could solve this easily by hand.

Answer by SpaceManSpiff for Find the direction

Fri, 11 Jul 2014 09:08:08 -0500

To solve the system of equations $u_{1}^{2}+u_{2}^{2}=1$ and $2u_{1}+u_{2}=1$. think of them as a circle $x^{2}+y^{2}=1$ and a line $y=-2x+1$ and plug in the equation of the line for $y$ in the circle equation. This leaves you with $$(-2x+1)^2+x^2=1$$. When you foil out the polynomial you get $$5x^2-4x=0$$ $$\longrightarrow x(5x-4)=0$$ $x=0$ and $x=\frac{4}{5}$ are the solutions. The solutions to this equation provide the intersections of the line and circle described earlier which you can use to construct the appropriate unit vectors.