Find r(t): Given Initial Conditions

mamma_mia66
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Homework Statement


Given \vec{}r"(t)= 6i-4cos (2t)j+ 9e3tk,
r'\vec{} (0)= 4i +3k and r\vec{}(0)=j+k, find r\vec{}(t).

Homework Equations





The Attempt at a Solution



I will appreciate any ideas how to start this problem. Thank you.
 
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mamma_mia66 said:

Homework Statement


Given \vec{}r"(t)= 6i-4cos (2t)j+ 9e3tk,
r'\vec{} (0)= 4i +3k and r\vec{}(0)=j+k, find r\vec{}(t).

Homework Equations





The Attempt at a Solution



I will appreciate any ideas how to start this problem. Thank you.
If r(t) = x(t)i + y(t)j + z(t)k, then r''(t) = x''(t)i + y''(t)j + z''(t)k.

How you would you normally solve x''(t)=6 for x(t), given x'(0) and x(0) ?

What is troubling you?
 
I get it now. I need to integrate the given. Thank you.
 
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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