Vector valued functions and normal vectors

ahmetbaba
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Homework Statement


Find T(t), N(t), aT, and aN

r(t)= <wt-sinwt,1-coswt> t=t0


Homework Equations





The Attempt at a Solution

 
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You are apparently new here. To receive help on a problem such as this you are expected to fill in parts 2 and 3 to show what you know and what you have tried, and where you are stuck. Nobody here will just work the problem for you.
 
You need to make some attempt at the problem in order to receive assistance here. Your textbook should have formulas for finding the (unit?) tangent and normal for a general curve...
 
Yeah, I'm new, so let's see.

r(t0)= (wt0-sinwt0)i + ( 1-coswt0)j

v(t0)= r(t0)'. Well this is the first place I'm stuck, I don't know what w stands for. The instructor didn't get this far, however the h-w is due, and this is one of the questions.

a(t0)= v(t0)'

T(t0)= r'(t0)/[r'(t0)]

N(t0)= T'(t)/[T'(t)]

aT= a.T=v.a/[v]

aN=a.N=[v x a]/[v]

So,

If I could do the first two derivatives of this problem, I should be fine. oh, and the [] stand for magnitude(length) of the vectors.
 
"w" is just a constant (angular frequency), so treat it like one and find the derivatives of r(t)...
 
ahmetbaba said:
Yeah, I'm new, so let's see.

r(t0)= (wt0-sinwt0)i + ( 1-coswt0)j

v(t0)= r(t0)'. Well this is the first place I'm stuck, I don't know what w stands for. The instructor didn't get this far, however the h-w is due, and this is one of the questions.
This is mathematics, not physics! For the mathematics, it doesn't matter what the letters stand for. The variable is t0 and that is all you need to know to differentiate. "w" (its really \omega, the Greek letter "omega", commonly used for "angular frequency" as gabbagabbahey said, but, again, you don't need to know that to solve this problem.) is just a constant.

a(t0)= v(t0)'

T(t0)= r'(t0)/[r'(t0)]

N(t0)= T'(t)/[T'(t)]

aT= a.T=v.a/[v]

aN=a.N=[v x a]/[v]

So,

If I could do the first two derivatives of this problem, I should be fine. oh, and the [] stand for magnitude(length) of the vectors.

Do you know how to find the derivatives, with respect to t, of "at", "sin(bt)", and "cos(bt)"? That's all you need.
 
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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