Multivariable Calculus Question of the Day

1. $\langle 4, 2 \rangle$ (correct answer)
2. $\langle 2, 4 \rangle$
3. $\langle \frac{4}{\sqrt{5}}, \frac{2}{\sqrt{5}} \rangle$
4. $\langle \frac{20}{3}, -\frac{10}{3} \rangle$

Explanation: Let $\nabla f(1,2) = \langle a, b \rangle$. The direction from $P$ to $Q$ is $\vec{v}_1 = Q-P = \langle 2, 1 \rangle$. The unit vector is $\vec{u}_1 = \frac{1}{\sqrt{2^2+1^2}} \langle 2, 1 \rangle = \frac{1}{\sqrt{5}} \langle 2, 1 \rangle$. The directional derivative is $D_{\vec{u}_1}f = \nabla f \cdot \vec{u}_1 = \langle a, b \rangle \cdot \frac{1}{\sqrt{5}} \langle 2, 1 \rangle = \frac{2a+b}{\sqrt{5}}$. We are given this is $2\sqrt{5}$, so $\frac{2a+b}{\sqrt{5}} = 2\sqrt{5}$, which implies $2a+b=10$. The direction from $P$ to $R$ is $\vec{v}_2 = R-P = \langle -1, 2 \rangle$. The unit vector is $\vec{u}_2 = \frac{1}{\sqrt{(-1)^2+2^2}} \langle -1, 2 \rangle = \frac{1}{\sqrt{5}} \langle -1, 2 \rangle$. The directional derivative is $D_{\vec{u}_2}f = \nabla f \cdot \vec{u}_2 = \langle a, b \rangle \cdot \frac{1}{\sqrt{5}} \langle -1, 2 \rangle = \frac{-a+2b}{\sqrt{5}}$. We are given this is $0$, so $-a+2b=0$, which means $a=2b$. Substituting this into the first equation gives $2(2b)+b=10$, so $5b=10$, and $b=2$. Then $a=2(2)=4$. Thus, $\nabla f(1,2) = \langle 4, 2 \rangle$.