Parametric Equations describing curves

[RJLiberator]

Homework Statement

Homework Equations

The Attempt at a Solution

For part A) my answer was:[/B]
$\int_a^b \sqrt{(dx/dt)^2+(dy/dt)^2}dt$

The work I used for part A was based off this sites explanation: http://tutorial.math.lamar.edu/Classes/CalcII/ParaArcLength.aspx

For part B)
I simple took the derivatives with respect to t of the giving x and y functions then squared them and solved the integral from 0 to 1.
The answer I received was 3/2.
I have strong verification that this is correct.

Part C)
Part C is where I am stuck. When they say y=F(x) I usually connect a capital F with the antiderivative. Does this hold true?
How do we describe the curve using y=F(x)?

Part D seems rather straightforward once part C is understood.

Thank you for any help.

 

[HallsofIvy]

You are given $$x= (4/3)t^{3/2}$$ and $$y= t- (1/2)t^2$$ and you want to write y as a function of x.

Solve the first equation for t as a function of x and replace the t in the second equation with that.

 

[RJLiberator]

[(3/4)*x]^(2/3)=t

y = [(3/4)*x]^(2/3)-(1/2)([(3/4)*x]^(2/3))^2

This is writing y as a function of x, correct?

So now with this information I can plug in x = 0 and x = 1 to get the range of x?

And then I can use this equation to solve for part D? I assume.

Kind regards.

 

[LCKurtz]

[(3/4)*x]^(2/3)=t

y = [(3/4)*x]^(2/3)-(1/2)([(3/4)*x]^(2/3))^2

This is writing y as a function of x, correct?

Correct? Hard to say. If you would write it in Latex it would be readable.

So now with this information I can plug in x = 0 and x = 1 to get the range of x?

No. Those are the end point $t$ values. What are the corresponding $x$ values?

 

[RJLiberator]

Perhaps I need to use

(3x/4)^(2/3) = t and let 1 be 1 and 0 to find the range.

Thus the range would be from 0 to 4/3

 

[HallsofIvy]

Yes, the range is from x= 0 to x= 4/3. But I see no reason to solve for t to find that. You were given that $$x= (4/3)t^{3/2}$$ just set t equal to 0 and 1 in that.

 

[RJLiberator]

Excellent. Yes, thank you for confirming the result and showing the way.

:)