Question

The following parametric equations trace out a loop.

x= 7 - (4/2)t^2
y= (-4/6)t^3+4t+2

Find the t values at which the curve intersects itself:
t= +/?

What is the total area inside the loop?
Area = ?

Answer 1

The curve intersects itself at t=±2.

The total area inside the loop is 96/5 square units.

To find the t values at which the curve defined by the parametric equations intersects itself, we need to set the x and y equations equal to each other and solve for t:

`7-4/2t^2 =-4/6t^3 +4t+2`

Combine like terms and rearrange the equation to set it equal to zero:

`t^3-6t^2-12t-13=0`

Now, find the roots of this cubic equation. The roots are t=−2,1±
`√(5)`. However, we are only interested in the real values, so the curve intersects itself at t=−2.

To find the total area inside the loop, we can use the formula for the area enclosed by a parametric curve:

Area=
`\int\limits^ty(t)} \, dx`

In this case, the loop is traced from t=−2 to t=2. Substitute the given parametric equations into the integral and evaluate:

Area=
`\int\limits^2_2 {-4/6t^3+4t+2} \, dx`=96/5

Therefore, the total area inside the loop is 96/5 square units.