Find the directional derivative of ='false' f(x,y) = xy^2 in the direction ='false' \theta =...


Find the directional derivative of {eq}f(x,y) = xy^2 {/eq} in the direction {eq}\theta = \frac {\pi}{3} {/eq} at the point (2,4).

Answer and Explanation:

We begin by finding the partial derivatives and the gradient.

{eq}\displaystyle \begin{align*} f_x &= y^2 \\ f_y &= 2xy \\ \nabla f &= \langle y^2, 2xy \rangle \\ \nabla f(2, 4)= \langle 16, 16 \rangle \\ \end{align*} {/eq}

Next, we find a unit vector in the direction of the given angle.

{eq}\displaystyle \mathbf{u} = \langle \cos \frac{\pi}{3}, \sin \frac{\pi}{3} \rangle = \langle \frac{1}{2}, \frac{\sqrt{3}}{2} \rangle {/eq}

Finally, we use the formula to compute the directional derivative.

{eq}\displaystyle \begin{align*} D_\mathbf{u} f(2, 4) &= \langle 16, 16 \rangle \cdot \langle \frac{1}{2}, \frac{\sqrt{3}}{2} \rangle \\ &= 8(1 + \sqrt{3}) \end{align*} {/eq}

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