1. Nov 1, 2005

### JasonJo

1. calculate the length of the following paths:
f(t) = (7, t, t^2) over 1<t<3

the way i was taught in class to calculate the length of a curve is to integrate the magnitude of the velocity, but i cant seem to naturally integrate the magnitude of (0, 1, 2t)

similarly for:
f(t) = (cos(3t), sin(3t), 2t^(3/2)) 0<t<4

2. prove the bac-cab formula
aX(bXc) = b(a*c) - c(a*b)
where * is the dot product

what does b(a*c) mean?! how do i evaluate that?

i wrote aX(bXc) in terms of the component form, etc.

3. prove the length curve and curvature formulas.

5. prove aX(bXc) = (aXb)Xc + bX(aXc)

2. Nov 1, 2005

### HallsofIvy

Staff Emeritus
No reason why you can't! Personally, I dislike "physics" names for mathematical concepts: you integrate the magnitude of the tangent vector!
Yes, the tangent vector to f(t)= (7, t, t2) is j+ 2tk and its magnitude is $$\sqrt{1+ 4t^2}$$. The arclength you seek is
$$\int_1^3\sqrt{1+ 4t^2}dt$$. When I see that I immediately think "trig substitution". sin2u+ cos2u= 1 so, dividing by cos2 u, tan2u+ 1= sec2. Let 2t= tan u so that 2dt= sec2u du or dt= (1/2)sec2udu and $$\sqrt{1+ 4t^2}= \sqrt{1+ tan^2 u}= sec u$$. The integral becomes $$\frac{1}{2}\int sec^3 u du= \frac{1}{2}\frac{du}{cos^3 u}[/itex] (the limits of integration are u= arctan 2 and u= arctan 6. Multiply numerator and denominator by cos u to get [tex]\int \frac{cos u du}{cos^4 u}= \int \frac{cos u du}{(1- sin^2u)^2}$$. Make the change of variable x= sin u so that dx= cos u du and the integral becomes $$\int \frac{dx}{(1-x^2)^2}$$ which can be done by "partial fractions".
The tangent vector is -3sin(3t)i+ 3cos(3t)j+ 3t1/2k. The magnitude of that vector is $$\sqrt{9sin^2(3t)+ 9cos^2(3t)+ 9t}$$ which looks to me like $$3\sqrt{1+ t}$$. The arclength is $$3\int_0^4\sqrt{1+ t}dt$$. Don't you think "u= 1+t" would take care of that?
You just said * is the dot product so (a*c) is a number (scalar). b(a*c) is just that number times the vector b.
If $$a= a_1i+ a_2j+ a_3k$$, $$b= b_1i+ b_2j+ b_3k$$, and $$c= c_1i+ c_2j+ c_3k$$, then $$a*c= a_1c_1+ a_2c_2+ a_3c_3$$ and $$b(a*c)= (a_1c_1+ a_2c_2+ a_3c_3)b_1i+ (a_1c_1+ a_2c_2+ a_3c_3)b_2j+ (a_1c_1+ a_2c_2+ a_3c_3)b_3k$$
$$a*b= a_1b_1+ a_2b_2+ a_3b_3$$ so $$c(a*b)= (a_1b_1+ a_2b_2+ a_3b_3)c_1i+ (a_1b_1+ a_2b_2+ a_3b_3)c_2j+ (a_1b_1+ a_2b_2+ a_3b_3)c_3k$$
$$b(a*c)- c(a*b)= {(a_1c_1+ a_2c_2+ a_3c_3)b_1- (a*b= a_1b_1+ a_2b_2+ a_3b_3)c_1}i+ {(a_1c_1+ a_2c_2+ a_3c_3)b_2-(a_1b_1+ a_2b_2+ a_3b_3)c_2}j+ { (a_1c_1+ a_2c_2+ a_3c_3)b_3-(a_1b_1+ a_2b_2+ a_3b_3)c_3}k$$
Multiply that out and see if it is the same as you got for aX(bXC)
How you would do that depends on what theorems you have to work with.
Write out the components!

Last edited: Nov 1, 2005
3. Nov 1, 2005

### JasonJo

much thanks, possibly the most helpful homework post ever.

thanks!!!!!!!!!!!!!!!!!!!!!

4. Nov 1, 2005

### JasonJo

wait wait, one question:

prove the length of the path is equal to : int over a to b of sqrt(1+f'(x)) dx

using the fact that the integral of the magnitude of the velocity over the interval a to b is also equal to the length of the path

5. Nov 2, 2005

### benorin

hint: position(x)=<x,f(x)> and velocity(x)=$$\frac{d}{dx}$$position(x).

6. Nov 2, 2005

### amcavoy

You could also use two points (xn+1,yn+1) and (xn,yn) and use the Pythagorean Theorem to find the distance between them. Now sum those distances and let them get very small (in other words, write it as an integral). The last thing to do is use the Mean-Value Theorem to replace your ${y_n}^{2}$ with $\left(f'(x)\Delta x\right)^2$.