# Calculating Directional Derivative Given t

Find the derivative of $f(x, y, z) = xyz$ in the direction of the velocity vector of the helix

$r(t) = (\cos (6t) , \sin (6t) , 6t)$ at $t = \pi/6$.

I am not really sure how to solve this, since we are not explicitly given a point at which to evaluate $f_x = yz, f_y = xz, \text{ and } f_z = xy$.

• Let me know if my answer suffices. I just want to give you the best answer I can. – Mark Viola Mar 24 '15 at 4:24

I would assume that it wants the position defined by $r(\frac{\pi}{6})$. That would give you the point $x = \cos (\pi) = -1, y = \sin (\pi) = 0, z = \pi$. Then you can find the velocity vector $r'(\frac{\pi}{6})$, normalize it (divide by magnitude) and dot it with the gradient to find the directional derivative: $D_{r(\pi/6)}f = \nabla f \cdot \frac{r'(\pi/6)}{||r'(\pi/6)||}$.

• Oh okay I see. How do you know it wants the position defined by r(pi/6) though, and not that of v(pi/6)? It seems ambiguous to me. – Robert Mar 24 '15 at 4:00
• It's asking for the directional derivative with respect to the velocity when $t = \pi/6$, so it would make sense that we would be analyzing it at the point along the position vector. If you picture the helix and the vector "attached" to that point it makes intuitive sense. Plugging the components of $v$ into $f$ wouldn't really give you a meaningful point. – Patrick Mar 24 '15 at 4:06

Given the position vector $\vec r(t)=\hat x \cos(6t)+\hat y \sin(6t)+\hat z 6t$, the velocity $\vec v(t)$ is

\begin{align} \frac{d\vec r(t)}{dt} &= \hat x x'(t)+\hat y y'(t) +\hat z z'(t) \\ & = -6 \sin(6t)\hat x+6 \cos(6t) \hat y +6\hat z \end{align}

At $t=\frac{\pi}{6}$, the velocity is $\vec v(\frac{\pi}{6})=-6\hat y+6 \hat z$, for which the unit vector $\hat v$ is $\frac{\sqrt{2}}{2}(-\hat y+\hat z)$.

Now, the gradient at $t=\frac{\pi}{6}$ is the gradient at $x=-1$, $y=0$, and $z= \pi$. Thus, $\nabla f(-1,0,\pi)=-\hat y \pi$.

The directional derivative is then $\hat v(t=\frac{\pi}{6}) \cdot \nabla f(-1,0,\pi) =\frac{\sqrt{2}\pi}{2}$.