Question

Find the directional derivative of f(x,y,z)=2z2x+y3f(x,y,z)=2z2x+y3 at the point (−1,4,3)(−1,4,3) in the direction of the vector 15–√i+25–√j15i+25j. (Use symbolic notation and fractions where needed.)

Answer 1

$f(x,y,z)=2z^2x+y^3$

$f$ has gradient

$\nabla f(x,y,z)=2z^2\,\vec\imath+3y^2\,\vec\jmath+4xz\,\vec k$

which at the point (-1, 4, 3) has a value of

$\nabla f(-1,4,3)=18\,\vec\imath+48\,\vec\jmath-12\,\vec k$

I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say $\vec u=15\,\vec\imath+25\,\vec\jmath$, in which case we have

$\|\vec u\|=\sqrt{15^2+25^2}=5\sqrt{34}$

Then the derivative of $f$ at (-1, 4, 3) in the direction of $\vec u$ is

$D_{\vec u}f(-1,4,3)=\nabla f(-1,4,3)\cdot\dfrac{\vec u}{\|\vec u\|}=\boxed{\dfrac{294}{\sqrt{34}}}$