Figure 4 shows a sketch of part of the curve C with parametric equations $x = 3 an heta, \, y = 4 ext{ cos}^2 heta, \, 0 \leq \theta \leq \frac{\pi}{2}$ The point P lies on C and has coordinates (3, 2). The line l is the normal to C at P. The normal cuts the x-axis at the point Q. (a) Find the x coordinate of the point Q. The finite region S, shown shaded in Figure 4, is bounded by the curve C, the x-axis, the y-axis and the line l. This shaded region is rotated $2\pi$ radians about the x-axis to form a solid of revolution. (b) Find the exact value of the volume of this solid of revolution, giving your answer in the form $p x^2 + q x^2$, where p and q are rational numbers to be determined. [You may use the formula $V = \frac{1}{3}\pi r^2 h$ for the volume of a cone.]

A-Level Edexcel Maths Pure Trigonometric Functions: Figure 4 shows a sketch of part

Figure 4 shows a sketch of part of the curve C with parametric equations $x = 3 an heta, \, y = 4 ext{ cos}^2 heta, \, 0 \leq \theta \leq \frac{\pi}{2}$ The point P lies on C and has coordinates (3, 2) - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 7

Question 5

$Figure-4-shows-a-sketch-of-part-of-the-curve-C-with-parametric-equations--$x-=-3--an--heta,-\,-y-=-4--ext{-cos}^2--heta,-\,-0-\leq-\theta-\leq-\frac{\pi}{2}$--The-point-P-lies-on-C-and-has-coordinates-(3,-2)-Edexcel-A-Level Maths Pure-Question 5-2014-Paper 7.png$

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

Worked Solution & Example Answer:Figure 4 shows a sketch of part of the curve C with parametric equations $x = 3 an heta, \, y = 4 ext{ cos}^2 heta, \, 0 \leq \theta \leq \frac{\pi}{2}$ The point P lies on C and has coordinates (3, 2) - Edexcel - A-Level Maths Pure - Question 5 - 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 where the normal to the curve at point P intersects the x-axis, we first need to find the slope of the curve C at P.

Parametric Derivative: The parametric equations are given by:
- $x = 3\tan\theta$ , and
- $y = 4\cos^2\theta$ .
We calculate the derivatives:

$\frac{dx}{d\theta} = 3\sec^2\theta,$
$\frac{dy}{d\theta} = -8\cos\theta\sin\theta.$
Slope of the Curve: The slope of the curve C at P is found by:

$m = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-8\cos\theta\sin\theta}{3\sec^2\theta} = \frac{-8\cos^3\theta}{3}.$

Substituting the coordinates for P (3, 2) to find \theta:
- From $x = 3\tan\theta$ , we have $\tan\theta = 1 \Rightarrow \theta = \frac{\pi}{4}$ .
- Then substituting into $y = 4\cos^2\theta$ : $y = 2$ , confirms $\theta = \frac{\pi}{4}$ .
Evaluate the slope: Therefore, the slope at P is:

$m = \frac{-8\left(\frac{\sqrt{2}}{2}\right)^3}{3} = -\frac{8}{3\sqrt{2}}.$
Equation of the Normal: The slope of the normal line l is the negative reciprocal:

$m_{l} = \frac{3\sqrt{2}}{8}.$

The equation of the normal line using point-slope form is:

$y - 2 = \frac{3\sqrt{2}}{8}(x - 3).$
Where the Normal Cuts the x-Axis: To find point Q, set $y = 0$ :

$0 - 2 = \frac{3\sqrt{2}}{8}(x - 3) \Rightarrow -2 = \frac{3\sqrt{2}}{8}(x - 3).$

Solving for x gives:

$x - 3 = -\frac{16}{3\sqrt{2}} \Rightarrow x = 3 - \frac{16}{3\sqrt{2}} = \frac{9\sqrt{2} - 16}{3\sqrt{2}}.$
Final Answer: Thus the x-coordinate of point Q is:

$x_Q = \frac{9\sqrt{2}-16}{3\sqrt{2}}$ .

Step 2

Find the exact value of the volume of revolution.

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Answer

To find the volume of the solid of revolution generated by rotating region S about the x-axis, we utilize the method of cylindrical shells or disk/washer method.

Setting up the Volume Integral: The formula for the volume when revolving about the x-axis is given by:

$V = \pi \int_{a}^{b} y^2 \, dx$

where $y$ is expressed as a function of $x$ . Substituting our parametric equations, we have:

$V = \pi \int_0^3 (4\cos^2\theta)^2 \cdot \frac{dx}{d\theta} \cdot d\theta$

with limits of integration determined by the values of \theta that correspond to x from 0 to 3.
Change of Variables: From our earlier work, we found:
- For $x = 3\tan\theta$ , we have $dx = 3 \sec^2 \theta \, d\theta$ which makes,
$V = \pi \int_0^{\frac{\pi}{4}} (4\cos^2\theta)^2 (3\sec^2\theta) d\theta$

Evaluating the volume integral:

$= \pi \int_0^{\frac{\pi}{4}} 16 \cos^4\theta (3\sec^2\theta) d\theta$

$= 48\pi\int_0^{\frac{\pi}{4}} \cos^4\theta d\theta = 48\pi\left[\frac{3\pi}{16} - \frac{1}{8}\right].$
Final Calculation: The total volume thus simplifies to:

$V = \frac{9\pi}{2},$

leading to parameters p and q as determined in the problem statement.

Figure 4 shows a sketch of part of the curve C with parametric equations $x = 3 an heta, \, y = 4 ext{ cos}^2 heta, \, 0 \leq \theta \leq \frac{\pi}{2}$ The point P lies on C and has coordinates (3, 2) - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 7

Worked Solution & Example Answer:Figure 4 shows a sketch of part of the curve C with parametric equations $x = 3 an heta, \, y = 4 ext{ cos}^2 heta, \, 0 \leq \theta \leq \frac{\pi}{2}$ The point P lies on C and has coordinates (3, 2) - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 7

Find the x coordinate of the point Q.

Find the exact value of the volume of revolution.

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