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Figure 4 shows a sketch of part of the curve C with parametric equations $x = 3 an heta, \quad y = 4 ext{cos}^2 heta + 2, \quad 0 \leq \theta < \frac{\pi}{2}$ The point P lies on C and has coordinates (3, 2) - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 7

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Figure-4-shows-a-sketch-of-part-of-the-curve-C-with-parametric-equations--$x-=-3--an--heta,-\quad-y-=-4--ext{cos}^2--heta-+-2,-\quad-0-\leq-\theta-<-\frac{\pi}{2}$--The-point-P-lies-on-C-and-has-coordinates-(3,-2)-Edexcel-A-Level Maths Pure-Question 6-2014-Paper 7.png

Figure 4 shows a sketch of part of the curve C with parametric equations $x = 3 an heta, \quad y = 4 ext{cos}^2 heta + 2, \quad 0 \leq \theta < \frac{\pi}{2}$ ... show full transcript

Worked Solution & Example Answer:Figure 4 shows a sketch of part of the curve C with parametric equations $x = 3 an heta, \quad y = 4 ext{cos}^2 heta + 2, \quad 0 \leq \theta < \frac{\pi}{2}$ The point P lies on C and has coordinates (3, 2) - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 7

Step 1

Find the x coordinate of the point Q.

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Answer

To find the x-coordinate of point Q, we first need to calculate the slope of the normal line l at point P. The parametric equations give us:

dydθ=4cos(2θ)anddxdθ=3sec2(θ)\frac{dy}{d\theta} = 4 \cdot \text{cos}(2\theta) \quad \text{and} \quad \frac{dx}{d\theta} = 3 \sec^2(\theta)

Using the chain rule, we find the slope of the tangent line at P:

m(T)=dy/dθdx/dθ=4cos(2θ)3sec2(θ)m(T) = \frac{dy/d\theta}{dx/d\theta} = \frac{4 \cdot \text{cos}(2\theta)}{3 \sec^2(\theta)}

At point P when ( \theta = \frac{\pi}{4} ), substituting gives:

m(T)=4cos(π2)3sec2(π4)=0(since cos(π2) = 0)m(T) = \frac{4 \cdot \text{cos}(\frac{\pi}{2})}{3 \sec^2(\frac{\pi}{4})} = 0 \quad \text{(since cos($\frac{\pi}{2}$) = 0)}

Thus, the slope of the normal line, which is perpendicular to the tangent, will be undefined as well. The equation for the normal line will thus be vertical through P.

Since P has coordinates (3, 2), the line will intersect the x-axis at Q:

Q=(3,0)Q = (3, 0)

The x-coordinate of point Q is 3.

Step 2

Find the exact value of the volume of this solid of revolution.

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Answer

To compute the volume of the solid of revolution formed by rotating the region S around the x-axis, we use the disk method.

The radius of a typical disk at position x is given by the y-coordinate of the curve. We can find the volume V using the integral:

V=0xπ(function)2dxV = \int_{0}^{x} \pi (\text{function})^2 \, dx

Using the parametric equations, we express the y-values:

y=4cos2(θ)+2y = 4 \cos^2(\theta) + 2

From the relation x=3tanθx = 3\tan\theta, we have:

θ=arctan(x3)\theta = \arctan\left(\frac{x}{3}\right)

Substituting back into the volume integral leads to:

V=03π(4cos2(θ)+2)2dxV = \int_{0}^{3} \pi (4 \cos^2(\theta) + 2)^2 \, dx

Solving this integral, while applying the change of variables as required, ultimately gives:

V=929+6r2V = \frac{92}{9} + 6r^2

where ρ\rho and qq are rational numbers defined in the prompt.

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