How do I find the speed of an object with a given position function?

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Homework Statement


Find the speed of an object with r(t)=e^(t)<cos(t), sin(t), 1>.

Homework Equations


None.[/B]

The Attempt at a Solution


v(t)=e^(t)<cos(t)-sin(t), cos(t)+sin(t), 1>
I know that the speed is abs(v(t)) and the answer in the book is sqrt(3)*e^(t). So how do I find the speed?
 
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Math10 said:

Homework Statement


Find the speed of an object with r(t)=e^(t)<cos(t), sin(t), 1>.

Homework Equations


None.[/B]

The Attempt at a Solution


v(t)=e^(t)<cos(t)-sin(t), cos(t)+sin(t), 1>
I know that the speed is abs(v(t)) and the answer in the book is sqrt(3)*e^(t). So how do I find the speed?
The speed is the magnitude of your vector for v(t). How do you usually find the magnitude of a vector?
 
Never mind, I got it! Thanks for helping.
 
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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