A curve has parametric equations x = at² , y =
2at. Find the area bounded by the curve, the x-
axis, and the ordinates at t = 1 and t = 2
step plz
To find the area bounded by the curve, the x-axis, and the ordinates at t = 1 and t = 2, we can use the method of integration. Here are the steps:
Step 1: Determine the limits of integration:
Since we are given that t varies from 1 to 2, our limits of integration will be from t = 1 to t = 2.
Step 2: Find the equation for the x-coordinate of the curve:
Given that x = at², we can substitute t = 1 and t = 2 into the equation to find the x-coordinates at those values.
At t = 1: x₁ = a(1)² = a
At t = 2: x₂ = a(2)² = 4a
Step 3: Find the equation for the y-coordinate of the curve:
Given that y = 2at, we can substitute t = 1 and t = 2 into the equation to find the y-coordinates at those values.
At t = 1: y₁ = 2a(1) = 2a
At t = 2: y₂ = 2a(2) = 4a
Step 4: Write the integral representing the area:
The area bounded by the curve, x-axis, and the ordinates is given by the integral of y with respect to x from x = x₁ to x = x₂. In this case, the integral can be written as:
Area = ∫[x₁, x₂] y dx
Step 5: Convert the integral to be expressed in terms of t:
Since we want to express the integral in terms of t, we need to substitute the equation for x in terms of t into the integral.
Area = ∫[t=1, t=2] (2at) (dx/dt) dt
Step 6: Find dx/dt:
Differentiating x = at² with respect to t gives:
dx/dt = d(at²)/dt = 2at
Step 7: Evaluate the integral:
Now we can substitute all the expressions we found into the integral and evaluate it.
Area = ∫[t=1, t=2] (2at) (2at) dt
= 4a² ∫[t=1, t=2] t² dt
= 4a² [t³/3] [t=1, t=2]
= 4a² [(2³/3) - (1³/3)]
= 4a² (8/3 - 1/3)
= 4a² (7/3)
= (28/3) a²
Therefore, the area bounded by the curve, the x-axis, and the ordinates at t = 1 and t = 2 is (28/3) a².
To find the area bounded by the curve, the x-axis, and the ordinates at t = 1 and t = 2, we can use the formula for the area under a curve.
The first step is to find the limits of integration. In this case, the limits are t = 1 and t = 2, as given in the question.
Next, we need to find the equation for the curve in terms of y. Since we have the parametric equations x = at² and y = 2at, we can solve the first equation for t and substitute it into the second equation to get y in terms of x:
x = at² (solve for t)
t = sqrt(x/a)
Substituting t into the equation y = 2at, we get:
y = 2a(sqrt(x/a))
= 2sqrt(ax)
Now, we have the equation y = 2sqrt(ax) for the curve in terms of y.
To find the area under the curve, we can integrate this equation with respect to x from x = 0 to x = a:
A = ∫[0,a] 2sqrt(ax) dx
Integrating, we get:
A = [4/3 * a^(3/2)] |[0,a]
Evaluating this integral from 0 to a, we get:
A = 4/3 * a^(3/2)
So, the area bounded by the curve, the x-axis, and the ordinates at t = 1 and t = 2 is 4/3 * a^(3/2).
The curve is just a parabola:
x = y^2/4a
So, the area is just
A = ∫[1,2] y(t) dx(t)
dx = 2at dt
A = ∫[1,2] 2at * 2at dt
= ∫[1,2] 4a^2t^2 dt
= 4/3 a^2 t^3 [1,2]
= 4/3 a^2 (8-1)
= 28/3 a^2
To check, express y as a function of x:
A = ∫[a,4a] y dx
= ∫[a,4a] 2√(ax) dx
= 2√a (2/3 x^(3/2)) [a,4a]
= 4/3√a (8a√a - a√a)
= 4/3 √a * 7a√a
= 28/3 a^2