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The parametric equations of a curve are $x = 6 \sin^{2} t, y = 2 \sin 2 t + 3 \cos 2 t$ for $0 ⩽ t < π$ . The curve crosses the $x$ -axis at the points $B$ and $D$ and the stationary points are $A$ and $C,$ as shown in the diagram.

Find the value of the gradient of the curve at $B .$

Find the values of $t$ at $A$ and $C,$ giving each answer correct to $3$ decimal places.

Accepted Answer

Given, the parametric equations of a curve are $x = 6 \sin^{2} t, y = 2 \sin 2 t + 3 \cos 2 t$ for $0 \leq t \leq π$ .

And, the given graph of the curve is

It is given that $D$ , and $A$ are the stationary points, we need to find the values of $C,$ at $A$ , and $C$ .

We know that when a curve is given in parametric form, in terms of the parameter $t$ we can use the chain rule to find $\frac{d y}{d x}$ in terms of $t$ .

Now, $\frac{d x}{d t} = \frac{d}{d t} (6 \sin^{2} t)$

$= 6 \frac{d}{d t} (\sin^{2} t)$

$= 6 \times 2 \sin t \times \frac{d}{d t} (\sin t)$

$= 12 \sin t \cos t . . . . (i)$ $[∵ \frac{d}{d t} (\sin t) = \cos t]$

Now, $\frac{d y}{d t} = \frac{d}{d t} (2 \sin 2 t + 3 \cos 2 t)$

$= 2 \frac{d}{d t} (\sin 2 t) + 3 \frac{d}{d t} (\cos 2 t)$

$= 2 \times (\cos 2 t) \times \frac{d}{d t} (2 t) + 3 \times (- \sin 2 t) \times \frac{d}{d t} (2 t)$ $[∵ \frac{d}{d x} (\sin x) = \cos x, \frac{d}{d x} (\cos x) = - \sin x]$

$= 2 \times (\cos 2 t) \times 2 + 3 \times (- \sin 2 t) \times 2$ $[∵ \frac{d}{d x} (x^{n}) = n x^{n - 1}, n \in R]$

$= 4 \cos 2 t - 6 \sin 2 t . . . . (i i)$

Since $\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d x}{d t}$

$= (4 \cos 2 t - 6 \sin 2 t) \times (\frac{1}{12 \sin t \cos t})$ [Using $(i)$ , and $(i i)$ ]

$= \frac{(2 \cos 2 t - 3 \sin 2 t)}{6 \sin t \cos t}$

$= \frac{(2 \cos 2 t - 3 \sin 2 t)}{3 \sin 2 t}$

$= \frac{2}{3} \frac{\cos 2 t}{\sin 2 t} - 1$

$= \frac{2}{3} \cot 2 t - 1$

Hence, $\frac{d y}{d x} = \frac{2}{3} \cot 2 t - 1$ .

Now, for the stationary points we must have

$\frac{d y}{d x} = 0$

$\Rightarrow \frac{2}{3} \cot 2 t - 1 = 0$

$\Rightarrow \cot 2 t = \frac{3}{2} . . . (i i i)$

$\Rightarrow 2 t = \cot^{- 1} (1.5)$

$\Rightarrow 2 t = 0.588003$ radians

$\Rightarrow t = 0.294,$

We can see that the point $C$ lies in the fourth quadrant. So, from equation $(i i i)$ , we have

$\cot (2 π - 2 t) = - 1.5$

$\Rightarrow 2 π - 2 t = \cot^{- 1} (- 1.5)$

$\Rightarrow 2 π - 2 t = π - \cot^{- 1} (1.5)$

$\Rightarrow 2 π - 2 t = π - 0.588003$

$\Rightarrow 2 t = π + 0.588003$

$\Rightarrow 2 t = 3.14 + 0.588003$

$\Rightarrow 2 t = 3.72803$

$\Rightarrow t = \frac{3.72803}{2} = 1.864$

Hence, the required values of $t$ at $A$ , and $C$ are $0.294,$ and $1.864$ respectively.

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Differentiation, Exercise 9: END-OF-CHAPTER REVIEW EXERCISE 4

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Differentiation, Exercise 9: END-OF-CHAPTER REVIEW EXERCISE 4

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Differentiation, Exercise 9: END-OF-CHAPTER REVIEW EXERCISE 4 with Hints & Solutions

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