Find the area of the region enclosed by the parametric equation x = t^3−3t y = 6t^2.
The loop is formed when x=0, or t=±√3
So, the area is
∫[-√3,√3] y(t) dx/dt dt
= ∫[-√3,√3] (6t^2)(3t^2-3) dt = 144√3/5
Oh, finding the area, huh? Well, let's throw some math pies and see what flavors we get. To find the area enclosed by the parametric equation x = t^3 - 3t and y = 6t^2, we can start by finding the equation for the curve. By eliminating t from the equations, we find x = (y/6)^(3/2) - 3(y/6), which simplifies to x = (y^(3/2))/6 - y/2. Now, we can find the area by integrating this equation with respect to y over the appropriate bounds. Crunching those numbers may not make you laugh, but hey, at least you'll have the area in the end!
To find the area of the region enclosed by the parametric equation x = t^3 - 3t and y = 6t^2, we can use the formula for area in parametric equations.
The formula to find the area of a region defined by parametric equations x = f(t) and y = g(t) is:
A = ∫[a,b] | y * dx/dt | dt
In this case, we have x = t^3 - 3t and y = 6t^2. We need to find dx/dt.
Taking the derivative of x with respect to t, we get:
dx/dt = 3t^2 - 3
Now we can calculate the area:
A = ∫[a,b] | y * (3t^2 - 3) | dt
First, let's find the bounds of integration, a and b. We need to find the values of t where the curve intersects and encloses the region.
From the given parametric equations:
x = t^3 - 3t
y = 6t^2
To find the bounds of integration, we equate the x values:
t^3 - 3t = 0
Factoring out t, we get:
t(t^2 - 3) = 0
So, t = 0 or t^2 - 3 = 0.
Solving the second equation for t, we get:
t^2 = 3
Taking the square root, we have:
t = ±√3
Therefore, the bounds of integration are -√3 and √3.
Now we can calculate the area:
A = ∫[-√3, √3] | 6t^2 * (3t^2 - 3) | dt
Simplifying, we have:
A = 6 ∫[-√3, √3] | (3t^4 - 3t^2) | dt
Integrating term by term, we get:
A = 6 [ t^5/5 - t^3/3 ] |_−√3^√3
Evaluating the definite integral, we have:
A = 6 [ (√3^5/5 - √3^3/3) - (-√3^5/5 + √3^3/3) ]
Simplifying further, we get:
A = 6 (2√3^5/5 - 2√3^3/3)
Finally, calculating the area:
A = 6 * (2 * (3√3)/5 - 2 * √3)
A = 6 * (6√3/5 - 2√3)
A = 36√3/5 - 12√3/5
A = (36√3 - 12√3)/5
A = 24√3/5
Therefore, the area of the region enclosed by the parametric equations x = t^3 - 3t and y = 6t^2 is 24√3/5 square units.
To find the area of the region enclosed by a parametric equation, you can follow these steps:
1. Solve one of the parametric equations for the parameter variable. In this case, solve the second equation, y = 6t^2, for t: t = sqrt(y/6).
2. Substitute the expression for t back into the other parametric equation. In this case, substitute t = sqrt(y/6) into x = t^3 - 3t: x = (sqrt(y/6))^3 - 3(sqrt(y/6))^1 = (y^(3/2))/(6^(3/2)) - 3(y^(1/2))/(6^(1/2)).
3. Determine the range of y-values that the parametric equation covers. Since t = sqrt(y/6), we need to find the range of y when t varies. Since y = 6t^2, it is clear that y is always greater than or equal to 0.
4. Calculate the definite integral of the expression for x over the range of y-values. In this case, integrate x with respect to y, from 0 to a certain y-value that determines the enclosed region.
5. Evaluate the definite integral to find the area of the region enclosed by the parametric equation.
Now, let's calculate the area of the region enclosed by the given parametric equation.
Substituting t = sqrt(y/6) into x = t^3 - 3t, we have:
x = (sqrt(y/6))^3 - 3(sqrt(y/6))^1
x = (y^(3/2))/(6^(3/2)) - 3(y^(1/2))/(6^(1/2))
Next, we integrate x with respect to y over the range of y-values from 0 to a certain y-value that determines the enclosed region:
∫[(y^(3/2))/(6^(3/2)) - 3(y^(1/2))/(6^(1/2))] dy from 0 to y = b
Evaluating the integral, we get:
∫[(y^(3/2))/(6^(3/2)) - 3(y^(1/2))/(6^(1/2))] dy = [2(y^(5/2))/(5(6^(3/2))) - 6(y^(3/2))/(3(6^(1/2)))] evaluated from 0 to b
Simplifying further, we have:
[2(b^(5/2))/(5(6^(3/2))) - 6(b^(3/2))/(3(6^(1/2)))] - [2(0^(5/2))/(5(6^(3/2))) - 6(0^(3/2))/(3(6^(1/2)))]
Since the second term is zero, we are left with:
[2(b^(5/2))/(5(6^(3/2))) - 6(b^(3/2))/(3(6^(1/2)))]
This expression gives the area of the region enclosed by the given parametric equation in terms of the parameter variable y.