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Today, 20:30
Figure 6
Figure 6 shows a sketch of the curve 
C with parametric equations

x=8sin^(2)t quad y=2sin 2t+3sin t quad0 <= t <= (pi)/(2)
The region 
R, shown shaded in Figure 6, is bounded by 
C, the 
x-axis and the line with equation 
x=4
(a) Show that the area of 
R is given by

int_(0)^(a)(8-8cos 4t+48sin^(2)t cos t)dt
where 
a is a constant to be found.
(b) Hence, using algebraic integration, find the exact area of 
R.
(4)
Ndokumbirawo help apo

Today, 2020:3030\newlineFigure 66\newlineFigure 66 shows a sketch of the curve \newlineCC with parametric equations\newlinex=8sin2ty=2sin2t+3sint0tπ2x=8\sin^{2}t \quad y=2\sin 2t+3\sin t \quad0 \leq t \leq \frac{\pi}{2}\newlineThe region \newlineRR, shown shaded in Figure 66, is bounded by \newlineCC, the \newlinexx-axis and the line with equation \newlinex=4x=4\newline(a) Show that the area of \newlineRR is given by\newline0a(88cos4t+48sin2tcost)dt\int_{0}^{a}(8-8\cos 4t+48\sin^{2}t \cos t)dt\newlinewhere \newlineaa is a constant to be found.\newline(b) Hence, using algebraic integration, find the exact area of \newlineRR.\newline(44)\newlineNdokumbirawo help apo

Full solution

Q. Today, 2020:3030\newlineFigure 66\newlineFigure 66 shows a sketch of the curve \newlineCC with parametric equations\newlinex=8sin2ty=2sin2t+3sint0tπ2x=8\sin^{2}t \quad y=2\sin 2t+3\sin t \quad0 \leq t \leq \frac{\pi}{2}\newlineThe region \newlineRR, shown shaded in Figure 66, is bounded by \newlineCC, the \newlinexx-axis and the line with equation \newlinex=4x=4\newline(a) Show that the area of \newlineRR is given by\newline0a(88cos4t+48sin2tcost)dt\int_{0}^{a}(8-8\cos 4t+48\sin^{2}t \cos t)dt\newlinewhere \newlineaa is a constant to be found.\newline(b) Hence, using algebraic integration, find the exact area of \newlineRR.\newline(44)\newlineNdokumbirawo help apo
  1. Identify Equations and Bounds: Identify the parametric equations and the bounds of integration.\newlineThe parametric equations given are:\newlinex=8sin2(t)x = 8\sin^2(t)\newliney=2sin(2t)+3sin(t)y = 2\sin(2t) + 3\sin(t)\newlineThe bounds for tt are from 00 to π/2\pi/2.
  2. Intersection Point with Line: Determine the intersection point of the curve with the line x=4x=4.\newlineSet x=8sin2(t)x = 8\sin^2(t) equal to 44:\newline8sin2(t)=48\sin^2(t) = 4\newlinesin2(t)=12\sin^2(t) = \frac{1}{2}\newlinesin(t)=±12=±12\sin(t) = \pm\sqrt{\frac{1}{2}} = \pm\frac{1}{\sqrt{2}}\newlineSince 0tπ20 \leq t \leq \frac{\pi}{2}, sin(t)=12\sin(t) = \frac{1}{\sqrt{2}} at t=π4t = \frac{\pi}{4}.
  3. Calculate Area Integral: Calculate the area integral.\newlineThe area under the curve from t=0t=0 to t=π4t=\frac{\pi}{4} is given by the integral of ydxdtdty \frac{dx}{dt} dt.\newlinedxdt=16sin(t)cos(t)=8sin(2t)\frac{dx}{dt} = 16\sin(t)\cos(t) = 8\sin(2t) (using double angle identity)\newlineThe integral becomes:\newline0π4(2sin(2t)+3sin(t))8sin(2t)dt\int_{0}^{\frac{\pi}{4}} (2\sin(2t) + 3\sin(t)) \cdot 8\sin(2t) dt\newline=0π4(16sin2(2t)+24sin(2t)sin(t))dt= \int_{0}^{\frac{\pi}{4}} (16\sin^2(2t) + 24\sin(2t)\sin(t)) dt
  4. Simplify Integral: Simplify and solve the integral.\newlineUsing trigonometric identities:\newlinesin2(2t)=1cos(4t)2\sin^2(2t) = \frac{1 - \cos(4t)}{2}\newlinesin(2t)sin(t)=cos(t)cos(3t)2\sin(2t)\sin(t) = \frac{\cos(t) - \cos(3t)}{2}\newlineThe integral becomes:\newline0π/4(88cos(4t)+12cos(t)12cos(3t))dt\int_{0}^{\pi/4} (8 - 8\cos(4t) + 12\cos(t) - 12\cos(3t)) \, dt
  5. Evaluate Integral: Evaluate the integral.\newlineThe integral of cos(4t)\cos(4t) from 00 to π/4\pi/4 is 00.\newlineThe integral of cos(t)\cos(t) from 00 to π/4\pi/4 is sin(π/4)sin(0)=2/2\sin(\pi/4) - \sin(0) = \sqrt{2}/2.\newlineThe integral of cos(3t)\cos(3t) from 00 to π/4\pi/4 is 0011.\newlineThe area is:\newline0022\newline0033\newline0044

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