# Parametric Equation Integration

• Mar 27th 2008, 05:10 AM
Del
Parametric Equation Integration
Notice that the curve given by the parametric equations

x = 81 - t^2

y = t^3 - 25t

is symetric about the x-axis. (If t gives us the point (x,y), then -t will give (x,-y) ).

At which x value is the tangent to this curve horizontal?

The curve makes a loop around the x-axis. What is the total area of this loop?

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• Mar 27th 2008, 06:50 AM
wingless
$x = 81 - t^2$

$\frac{dx}{dt} = -2t$

$y = t^3 - 25t$

$\frac{dy}{dt} = 3t^2 - 25$

$\frac{dy}{dx} = \frac{dy}{dt}\cdot \frac{dt}{dx} = \frac{3t^2-25}{-2t}$

$y' = \frac{25-3t^2}{2t}$

Now set $y'=0$ (cause its slope is 0) and find its corresponding x coordinate.
• Mar 27th 2008, 07:01 AM
Del
**post deleted** Thanks for the help.
• Mar 27th 2008, 07:10 AM
CaptainBlack
Quote:

Originally Posted by Del
Notice that the curve given by the parametric equations

x = 81 - t^2

y = t^3 - 25t

is symetric about the x-axis. (If t gives us the point (x,y), then -t will give (x,-y) ).

At which x value is the tangent to this curve horizontal?

The curve makes a loop around the x-axis. What is the total area of this loop?

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The curve is horizontal when:

$\frac{dy}{dx}=0$,

differentiating the equation for $y$ in terms of $t$ by $x$ give:

$\frac{dy}{dx}=3t^2\frac{dt}{dx}-25\frac{dt}{dx}$

so at the point of interest either $\frac{dt}{dx}=0$ or $t=\sqrt{\frac{25}{3}}.$

I will leave it to you to show that the first of these does not occur, and to
find the value of $x$ corresponding to the latter.

RonL
• Mar 27th 2008, 08:29 AM
Del
Thanks guys, but what about the area of the loop about the x-axis?
• Mar 27th 2008, 09:44 AM
topsquark
Quote:

Originally Posted by Del
x = 81 - t^2

y = t^3 - 25t

is symetric about the x-axis. (If t gives us the point (x,y), then -t will give (x,-y) ).

Quote:

Originally Posted by Del
Thanks guys, but what about the area of the loop about the x-axis?

The relation is symmetric about the x-axis. So we need to know for what t values the loop is traced. So we know that the curve must meet itself on the x-axis, so we look for where that happens:
$t^3 - 25t = 0$

$t(t + 5)(t - 5) = 0$

So t = -5, 0, and 5. This corresponds to x values:
x = 56, 81, and 56 respectively.

So if we find the area between the curve and the x-axis over the [0, 5] t interval and double it, we will have the area of the loop.

Does this help?

-Dan
• Mar 27th 2008, 03:09 PM
Del
This is helpful, but I keep getting a negative value when I integrate. I got -2333.333, but this is incorrect.
• Mar 27th 2008, 03:55 PM
TheEmptySet
counter clock wise :)
Quote:

Originally Posted by Del
This is helpful, but I keep getting a negative value when I integrate. I got -2333.333, but this is incorrect.

The curve is being traversed in a clockwise manner NOT counter clockwise.

Green's Theorem states that the area is

$A=\int_{c}xdy=-\int_{c}ydx$ if c is positively oriented

so we also know that

$\int_{c}xdy=-\int_{-c}xdy$ where -c is going around the curve clockwise.

I hope this helps.