Today, 20:30Figure 6Figure 6 shows a sketch of the curve C with parametric equationsx=8sin2ty=2sin2t+3sint0≤t≤2πThe 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∫0a(8−8cos4t+48sin2tcost)dtwhere a is a constant to be found.(b) Hence, using algebraic integration, find the exact area of R.(4)Ndokumbirawo help apo
Q. Today, 20:30Figure 6Figure 6 shows a sketch of the curve C with parametric equationsx=8sin2ty=2sin2t+3sint0≤t≤2πThe 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∫0a(8−8cos4t+48sin2tcost)dtwhere a is a constant to be found.(b) Hence, using algebraic integration, find the exact area of R.(4)Ndokumbirawo help apo
Identify Equations and Bounds: Identify the parametric equations and the bounds of integration.The parametric equations given are:x=8sin2(t)y=2sin(2t)+3sin(t)The bounds for t are from 0 to π/2.
Intersection Point with Line: Determine the intersection point of the curve with the line x=4.Set x=8sin2(t) equal to 4:8sin2(t)=4sin2(t)=21sin(t)=±21=±21Since 0≤t≤2π, sin(t)=21 at t=4π.
Calculate Area Integral: Calculate the area integral.The area under the curve from t=0 to t=4π is given by the integral of ydtdxdt.dtdx=16sin(t)cos(t)=8sin(2t) (using double angle identity)The integral becomes:∫04π(2sin(2t)+3sin(t))⋅8sin(2t)dt=∫04π(16sin2(2t)+24sin(2t)sin(t))dt
Simplify Integral: Simplify and solve the integral.Using trigonometric identities:sin2(2t)=21−cos(4t)sin(2t)sin(t)=2cos(t)−cos(3t)The integral becomes:∫0π/4(8−8cos(4t)+12cos(t)−12cos(3t))dt
Evaluate Integral: Evaluate the integral.The integral of cos(4t) from 0 to π/4 is 0.The integral of cos(t) from 0 to π/4 is sin(π/4)−sin(0)=2/2.The integral of cos(3t) from 0 to π/4 is 01.The area is:020304
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