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Problem:

(a) Find the directional derivative of $w=x^2+y^2$ in the direction of the tangent vector to the spiral $(x, y) = (e^t \cos(t), 2e^t \sin(t))$, at the point defined by $t=0$. Done.

(b) Find $\frac{dw}{dt}$ along the spiral, at the same point. Done

(c) How are these rates of change related? This is the difficult part

Attempt at solution for (c):

The rate of change in (a) is the directional rate of change of $w$ in the direction of $(1, 2)$, while the rate of change in (b) is the rate of change of $w$ with respect to $t$. Both derivatives are at the same point and in the same direction, but the derivative in (b) changes faster because of the rate of change in $t$.

I feel that my solution lacks some insight. Would appreciate a hint.

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Hint: The following is from Peter Oliver's notes at the University of Minnesota;since he does such a good job explaining the connection,why should I knock myself out giving you my explanation? This should give you a deeper understanding of how the 2 derivatives are connected and allow you to reword your answer accordingly.

Oliver's full notes can be found here.

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.....direction. Sorry for cutting off the page.

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  • $\begingroup$ Thanks. But in part (a) we find that the directional derivative is $2/sqrt{5}$, while in part (b) it's 2. They are not the same, so it's still a question how they might be related. Calculations for part (a): (i) first I found the tangent vector to the spiral and set t to 0. (ii) I normalized the tangent vector, which came to $(1/sqrt{5}, 2/sqrt{5})$. (iii) the dot product of the gradient with the above vector came to $2/sqrt{5}$. $\endgroup$ – sequence Feb 13 '15 at 5:39
  • $\begingroup$ They are clearly NOT the same. But the positive gradient points along the direction of steepest ascent and the dot product yields the chain rule expression for the cylinder along the helical tangent path via parametrization. If it's still not clear,try drawing the graph either by hand or computer sketch. $\endgroup$ – Mathemagician1234 Feb 13 '15 at 6:40
  • $\begingroup$ The gradient part is clear of course, but the cylinder part is not yet, because I haven't yet covered the material pertaining to cylindrical tangent paths. What is clear to me, however, is that while in the first case we're having a simple directional derivative, in the second case we're talking about the same direction, yet the length of the vector is not unit (normalized). In the first case, we're getting the rate of change of height with respect to length, in the second case - height with respect to the parametric variable t. Please correct me if I'm wrong. $\endgroup$ – sequence Feb 15 '15 at 0:33

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