Describing the path of a heat seeking particle

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Homework Statement



Find the path of a heat-seeking particle placed at (4,3,10) with a temperature field

T(x,y,z) = 400 - 2x^2 - y^2 -4z^2

Homework Equations



Formulas for directional derivative and gradient.

The Attempt at a Solution



At any point in space (x,y,z), the particle must be moving in the direction which causes the greatest increase in T. This direction is given by the direction of

\nabla T = < -4x, -2y, -8z >.

I do not know how to continue from here.
 
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The velocity vector points in the direction of \nabla T and so must be some multiple, say "k", of it.

That means you must have dx/dt= -4kx, dy/dt= -2ky, and dz/dt= -8kz for some number k.
You also are given that x(0)= 4(0), y(0)= 3, z(0)= 10. You can solve each of those for x, y, and z in terms of kt. That gives parametric equations for the path with parameter s= kt.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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