The following parametric equations trace out a loop. x=6−3/2(t^2) y=−3/6(t^3)+3t+1 Find the t values at which the curve intersects itself: t=± ___ .What is the total area inside the loop? Area = ___

Answer 1

In the absence of parentheses I not sure how to read the equations.

This is my interpretation

x(t) = 6 - (3/2)t²

y(t) = (-3/6)t³ + 3t + 1 = -t³/2 + 3t + 1

If I got it wrong, I hope that you will be able to adapt my solution to the correct equations.

May I suggest that you draw the curve in the x-y plane -- it will make things clearer.

Finding the t values at which the curve intersects itself means

finding two values t₁ ≠ t₂ satisfying

x(t₁ ) = x(t₂) (1)

y(t₁ ) = y(t₂) (2)

Equation (1) gives

6 - (3/2)t₁² = 6 - (3/2)t₂²

Simplify into

t₁² = t₂² (3)

Equation (2) gives

-t₁³/2 + 3t₁ + 1 = -t₂³/2 + 3t₂ + 1

Simplify into

t₁ (-t₁²/2 + 3) = t₂ (-t₂²/2 + 3)

Substituting from (3)

t₁ (-t₂²/2 + 3) = t₂ (-t₂²/2 + 3)

If we assume that (-t₂²/2 + 3) ≠ 0

then we can divide by (-t₂²/2 + 3)

giving us the trivial solution

t₁ = t₂

Non-trivial solution is obtained from the other alternative

(-t₂²/2 + 3) = 0

t₂ = ±√6

When combined with (3) we have that one of t₁, t₂ is √6 and the other is -√6.

Or in terms of the notation of the problem statement

t = ±√6

That means, to trace the loop, t ranges from -√6 to +√6.

As t ranges from -√6 to 0, x ranges from -3 to 6, and then

as t ranges from 0 to +√6, x ranges back from 6 to -3.

Use the definition of x(t) to express t as a function of x:

t = ±√(4 - 2x/3) (4)

That means we can trace the loop by letting x range from -3 to 6 using

t = -√(4 - 2x/3) (5)

and then let x range from 6 to -3 using

t = +√(4 - 2x/3) (6)

The next step is to find the area inside the loop.

It is a straightforward integration, but messy, so let me try to be careful.

Still an error is possible.

If you have a picture of the loop, it looks like a fish facing the positive x-direction.

You can think of the area inside the loop in two ways.

One way is to see it as the area under the upper curve minus the area under the lower curve.

The first area corresponds to t ranging from -√6 to 0; the other corresponds to t ranging from +√6 to 0.

The other view is to let t range from -√6 to 0 and then from 0 to +√6.

In this view we add two integrals together.

Both views amount to the same thing, I will use the second.

Let me first introduce a little notation.

Let

f(t) = -t³/2 + 3t

That means,

y(t) = f(t) + 1

For the sake of integration, express y as a function of x.

Into the definition of y(t) substitute either (5) or (6) for t.

So we get two functions; we name them y₁ and y₂.

When using (5)

y₁(x) = f(-√(4 - 2x/3)) + 1 (7)

When using (6)

y₂(x) = f(+√(4 - 2x/3)) + 1 (8)

From the definition of f(t) notice that f(-t) = -f(t).

Therefore we can rewrite (7) and (8) as

y₁(x) = -f(√(4 - 2x/3)) + 1 (9)

y₂(x) = f(√(4 - 2x/3)) + 1 (10)

Denote

F(x) = ∫ f(√(4 - 2x/3)) dx

Then we can define the integrals for y₁ and y₂ by integrating (9) and (10)

G₁(x) = ∫ y₁(x) dx = -F(x) + x

G₂(x) = ∫ y₂(x) dx = F(x) + x

The area enclosed by the loop then can be written as the sum of two definite integrals

A = { G₁(6) - G₁(-3) } + { G₂(-3) - G₂(6) } =

{ [-F(6) + 6] - [-F(-3) + (-3) ] } + { [ F(-3) + (-3) ] - [ F(6) + 6 ] } =

-F(6) + F(-3) + F(-3) - F(6) =

-2(F(-3) - F(6)) (11)

No we actually evaluate the integral we need:

F(x) = ∫ f(√(4 - 2x/3)) dx =

∫ [- (4 - 2x/3)^3/2 / 2 + 3 (4 - 2x/3)^1/2 ] dx =

-(2/5)(-3/2)(4 - 2x/3)^5/2 / 2 + 3 (2/3) (-3/2) (4 - 2x/3)^3/2 =

(3/10) (4 - 2x/3)^5/2 - 3(4 - 2x/3)^3/2 =

(4 - 2x/3)^3/2 [ (3/10) (4 - 2x/3) - 3] =

(4 - 2x/3)^3/2(-18 - 2x)/10

So

F(-3) = (4 - 2 (-3)/3)^3/2 (-18 - 2(-3))/10 = -6^3/2(12/10)

F(6) = 0

Finally from (11)

A = -2(-6^3/2(12/10)) = 6^3/2(12/5) = 35.27