Parametric Equations with Trig sub and int by parts

[NastyAccident]

Homework Statement

x=cos^2(t)
y=cos(t)

(a) Find the distance traveled by a particle with position (x, y) as t varies in the given time interval.

(b) What is the length of the curve?

Homework Equations

Length of an Arc: integral of alpha to beta sqrt((dx/dt)^2+(dy/dt)^2)

The Attempt at a Solution

See attached PDF for full attempt as soon as it clears...

Originally, I attempted to find a factor to easily cancel the square root out. However, with this particular problem I couldn't. So, I took out my bazooka and used Trig sub for sqrt(4u^2+1) & Integration by parts (for sec^3(s))

My answer ends up being 12(-20+ln(3)-ln(7))... Which is negative plus it is way too high for (a)...

I integrated from 0 to pi (the length of one arc, which is the answer to b) and times it by six (there are a total of six arches).

Any help is greatly appreciated and will be thanked, I've been staring at this for about 2 hours now and I still can't figure out what is off.



NastyAccident

 

[Mark44]

The graph of these parametric equations looks exactly like the graph of x = y^2, where 0 <= x <= 1, and -1 <= y <= 1. When t = 0, you get the point (1, 1), and when t = pi/2, you get the point (0,0), and when t = pi, you get the point (1, -1). That's one complete cycle. As t increases beyond pi, the points on the parabola are retraced.

It might be simpler to work with x as a function of y, rather than x and y as functions of t.

 

[LCKurtz]

I didn't check past your u substitution, but you have a glaring error there. That is that you didn't change to t limits to your new u limits.

[edit adding a couple of comments]

Also, if your original interval is larger than (0,\pi) you will need to use |sin(t)| when you remove it from the square root. And for
\int\sqrt{4u^2+1}du

try the substitution 2u = sinh(w)

 

[NastyAccident]

The graph of these parametric equations looks exactly like the graph of x = y^2, where 0 <= x <= 1, and -1 <= y <= 1. When t = 0, you get the point (1, 1), and when t = pi/2, you get the point (0,0), and when t = pi, you get the point (1, -1). That's one complete cycle. As t increases beyond pi, the points on the parabola are retraced.

It might be simpler to work with x as a function of y, rather than x and y as functions of t.

Oh, how I wish I could do that! However, the prof and the TA said we would be penalized for not using Parametric Equations in finding our answer since this chapter emphasizes the importances of parametric equations.

I didn't check past your u substitution, but you have a glaring error there. That is that you didn't change to t limits to your new u limits.

[edit adding a couple of comments]

Also, if your original interval is larger than (0,\pi) you will need to use |sin(t)| when you remove it from the square root. And for
\int\sqrt{4u^2+1}du

try the substitution 2u = sinh(w)

Well, if you convert the u substitution back to t, then technically, I thought, you can use your original limits.

Now, is this after I substitute in for cos(t)?
\int\sqrt{4u^2+1}du

2u=sinh(w) => 2(cos(t))=sinh(w)?

Correct? (Just want to make sure about this..)



NastyAccident

 

[LCKurtz]

Oh, how I wish I could do that! However, the prof and the TA said we would be penalized for not using Parametric Equations in finding our answer since this chapter emphasizes the importances of parametric equations.

That wouldn't make any difference. Doing it the xy way you would immediately be confronted with an integral just like the one below.

Well, if you convert the u substitution back to t, then technically, I thought, you can use your original limits.

Well, you can do that; I didn't check all your work. It is much more efficient to carry the limits along through the substitution so you never have to back substitute when doing a definite integral.

Now, is this after I substitute in for cos(t)?
\int\sqrt{4u^2+1}du

2u=sinh(w) => 2(cos(t))=sinh(w)?

Correct? (Just want to make sure about this..)

Your u integral is correct once you put the proper limits on it, but no, that last equation isn't correct. This is a hyperbolic function substitution and there is no cos(t) in it. You are changing the variable from u to w. You need to compute du and use the hyperbolic identity:

\cosh^2{w} = 1 + \sinh^2{w}