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Given the following function

$$f(x, y, z) = 2x ^2 + 4y − z$$

and the parametric curve L given by

$$x = t,$$ $$y = t − 1,$$

$$z = t ^2 + 4t$$,

$$t ∈ [0, 5].$$

Let $(a, b, c)$ be the point of intersection between the level surface $f(x, y, z) = 0$ and the curve L, Find the directional derivative of f at $(a, b, c)$ in the direction of the tangent to L at $(a, b, c)$ (where the tangent vector is taken in the direction of increasing t)

What i tried

I know the formula for finding directional derivative. $D=gradf.u$ where $u$ is the unit vector

I first find the gradient vector of $f(x,y,z)$ which gives $<4x,4,-1>$ substituting $(a, b, c)$ gives a gradient vector of $<4a,4,-1>$ but im stuck from here onwards, as i cant releate my parametric equation to my gradient vector? And im unsure of how to find the unit vector $u$ so that i can use the formula above. Could anyone help me with this. Thanks for any help.

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  • $\begingroup$ Use \nabla for $\nabla$ $\endgroup$ – Amitai Yuval Sep 25 '14 at 14:46
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Some thoughts:

  • The point of intersection satisfies the equation: $$ f(x,y,z)|_{L} = 0 \implies 2 t^2 + 4(t-1) - (t^2+4t) = 0,$$ whose solution is $t=\pm2$, but we are restricted to $t \in (0,5)$, so $t^*= 2 $ is the final choice. This gives us: $$ P_1(a,b,c) = P_1(2,1,12)$$ as the required point of intersection.

  • The normal tangent vector of a parametrized curve given by $\gamma(t) = (x(t),y(t),z(t))$ is $$\mathbf{u} = \frac{\tilde{\mathbf{u}}}{|| \tilde{\mathbf{u}}||}, \quad \mathbf{\tilde{u}} = \frac{\mathrm{d}}{\mathrm{d}t} \gamma(t) = (1,1,2t+4).$$ Substituing now $t=2$ will give us the desired direction.

  • Finally use:

$$ D_\mathbf{u} f|_P = \mathbf{u} \cdot \nabla f|_P $$

Hope this helps!

Cheers.

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