Calc 3 directional derivative question

meadow
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The question asks:
Find the directional derivative of f (x, y, z) = z ln (x/y) at (1, 1, 2) toward the point (2, 2, 1).

What I did was find the distance between the two points to be the directional vector (i+j-k) and then I took the norm of the direction vector. so my unit vector = 1/sqrt(3) * u; then I found the gradient. From there, I found the scalar product of my unit vector and the gradient to get 0. Did I approach this problem right? Does that answer seem correct to you?
 
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Seems all right to me :smile:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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