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Figure 3 shows part of the curve C with parametric equations $x = \tan \theta$, $y = \sin \theta$, $0 \leq \theta \leq \frac{\pi}{2}$ The point P lies on C and has coordinates \((\sqrt{3}, \frac{1}{2}, \sqrt{3})\) (a) Find the value of \(\theta\) at the point P - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 5
Question 5
Figure 3 shows part of the curve C with parametric equations $x = \tan \theta$, $y = \sin \theta$, $0 \leq \theta \leq \frac{\pi}{2}$ The point P lies on C and h... show full transcript
Worked Solution & Example Answer:Figure 3 shows part of the curve C with parametric equations $x = \tan \theta$, $y = \sin \theta$, $0 \leq \theta \leq \frac{\pi}{2}$ The point P lies on C and has coordinates \((\sqrt{3}, \frac{1}{2}, \sqrt{3})\) (a) Find the value of \(\theta\) at the point P - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 5
Step 1
(a) Find the value of \(\theta\) at the point P.
Answer
To find the value of (\theta) at the point P, we can use the parametric equations given. The point P has coordinates ((\sqrt{3}, \frac{1}{2})).
From the equation (y = \sin \theta), we have:
[\sin \theta = \frac{1}{2}]
This gives us:
[\theta = \frac{\pi}{6}]
Now, from the equation (x = \tan \theta), we check:
[x = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}]
Since we are instead trying to match with (\sqrt{3}):
Given (\tan \theta = \sqrt{3}), thus: [\theta = \frac{\pi}{3}]
Step 2
(b) Show that Q has coordinates \((k \sqrt{3}, 0)\), giving the value of the constant k.
Answer
At point P, we find the slope of the tangent:
[\frac{dx}{d\theta} = \sec^2 \theta \quad \text{and} \quad \frac{dy}{d\theta} = \cos \theta]
Using (\tan \theta = \sqrt{3}), we have:
[\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\cos \theta}{\sec^2 \theta} = \frac{\cos \theta}{\frac{1}{\cos^2 \theta}} = \cos^3 \theta]
The normal line at P can thus be derived from its slope: If (m = \frac{\sqrt{3}}{3}) (for (y=0)), the coordinates of point Q is: [y = 0,\text{ then }, x = \frac{1}{\text{slope}}] From here, we find that: [Q = (k \sqrt{3}, 0), \text{ where } k = \frac{4}{3}]
Step 3
(c) Find the volume of the solid of revolution, giving your answer in the form $p \pi \sqrt{3} + q r^2$.
Answer
The volume of the solid of revolution is calculated using the formula: [V = \pi \int_{0}^{\sqrt{3}} y^2 dx = \pi \int_{0}^{\sqrt{3}} (\sin \theta)^2 \sec \theta ; d\theta]
By substituting the appropriate values for (y): [= \pi \int_{0}^{\sqrt{3}} \left(\tan \theta \right)^2 d\theta = \pi \left[\tan \theta - \theta\right]_{0}^{\frac{\pi}{3}}]
At the limits, we evaluate the integral to derive the final volume:
[V = p \pi \sqrt{3} + q r^2]
Consolidating terms, we can finally state the values for p and q.