Find Tangent Line to Path: x(t)=4costi-3sintj+5tk, t=\pi/3

rgalvan2
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Find an equation for the line tangent to the given path at the indicated value for the parameter.

x(t)=4costi-3sintj+5tk, t=\pi/3

So what I did here was take x'(t) and then plugged in \pi/3 after that to get an equation containing i, j, and k.
x'(t)= -4sinti-3costj+5k
x'(\pi/3)=-4\sqrt{3}/2i-3/2j+5k
Am I doing this right? Calc III is confusing me so much. Any help would be appreciated. This is due Thursday. Thanks!
 
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Looks all right so far.
 
Alright sweet thanks a lot!
 

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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