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General Solution of sin theta = sin alpha

Find the equa...

Question

Find the equations of the tangent and the normal to the following curves at the indicated points.

(i) x = θ + sin θ, y = 1 + cos θ at θ = π/2
(ii) $x = \frac{2 a t^{2}}{1 + t^{2}}, y = \frac{2 a t^{3}}{1 + t^{2}} at t = 1 / 2$
(iii) x = at², y = 2at at t = 1
(iv) x = a sec t, y = b tan t at t
(v) x = a (θ + sin θ), y = a(1 − cos θ) at θ

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Solution

(i)
$x = θ + \sin θ and y = 1 + \cos θ \frac{d x}{d θ} = 1 + \cos θ and \frac{d y}{d θ} = - \sin θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{- \sin θ}{1 + \cos θ} Slope of tangent, m = {(\frac{d y}{d x})}_{θ = \frac{π}{2}} = \frac{- \sin \frac{π}{2}}{1 + \cos \frac{π}{2}} = \frac{- 1}{1 + 0} =-1 Now, (x_{1}, y_{1}) = (\frac{π}{2} + \sin \frac{π}{2}, 1 + \cos \frac{π}{2}) = (\frac{π}{2} + 1, 1) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 1 = - 1 (x - \frac{π}{2} - 1) \Rightarrow 2 y - 2 = - 2 x + π + 2 \Rightarrow 2 x + 2 y - π - 4 = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - 1 = 1 (x - \frac{π}{2} - 1) \Rightarrow 2 y - 2 = 2 x - π - 2 \Rightarrow 2 x - 2 y = π$

(ii) Equation of tangent:
$x = \frac{2 a t^{2}}{1 + t^{2}} and y = \frac{2 a t^{3}}{1 + t^{2}} \frac{d x}{d t} = \frac{(1 + t^{2}) (4 a t) - 2 a t^{2} (2 t)}{{(1 + t^{2})}^{2}} = \frac{4 a t}{{(1 + t^{2})}^{2}} and \frac{d y}{d t} = \frac{(1 + t^{2}) 6 a t^{2} - 2 a t^{3} (2 t)}{{(1 + t^{2})}^{2}} = \frac{6 a t^{2} + 2 a t^{4}}{{(1 + t^{2})}^{2}} \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{6 a t^{2} + 2 a t^{4}}{{(1 + t^{2})}^{2}}}{\frac{4 a t}{{(1 + t^{2})}^{2}}} = \frac{6 a t^{2} + 2 a t^{4}}{4 a t} Slope of tangent, m = {(\frac{d y}{d x})}_{t = \frac{1}{2}} = \frac{\frac{3 a}{2} + \frac{a}{8}}{2 a} = \frac{\frac{12 a + a}{8}}{2 a} = \frac{13 a}{8} × \frac{1}{2 a} = \frac{13}{16} Now, (x_{1}, y_{1}) = (\frac{2 a t^{2}}{1 + t^{2}}, \frac{2 a t^{3}}{1 + t^{2}}) = (\frac{\frac{a}{2}}{1 + \frac{1}{4}}, \frac{\frac{a}{4}}{1 + \frac{1}{4}}) = (\frac{\frac{a}{2}}{\frac{5}{4}}, \frac{\frac{a}{4}}{\frac{5}{4}}) = (\frac{2 a}{5}, \frac{a}{5}) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - \frac{a}{5} = \frac{13}{16} (x - \frac{2 a}{5}) \Rightarrow \frac{5 y - a}{5} = \frac{13}{16} (\frac{5 x - 2 a}{5}) \Rightarrow 5 y - a = \frac{13}{16} (5 x - 2 a) \Rightarrow 80 y - 16 a = 65 x - 26 a \Rightarrow 65 x - 80 y - 10 a = 0 \Rightarrow 13 x - 16 y - 2 a = 0$

$Equation of normal is, y - y_{1} = \frac{- 1}{m} (x - x_{1}) \Rightarrow y - \frac{a}{5} = \frac{- 16}{13} (x - \frac{2 a}{5}) \Rightarrow \frac{5 y - a}{5} = \frac{- 16}{13} (\frac{5 x - 2 a}{5}) \Rightarrow 5 y - a = \frac{- 16}{13} (5 x - 2 a) \Rightarrow 65 y - 13 a = - 80 x + 32 a \Rightarrow 80 x + 65 y - 45 a = 0 \Rightarrow 16 x + 13 y - 9 a = 0$

(iii)
$x = a t^{2} and y = 2 a t \frac{d x}{d t} = 2 a t and \frac{d y}{d t} = 2 a ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 a}{2 a t} = \frac{1}{t} Slope of tangent, m = {(\frac{d y}{d x})}_{t = 1} = \frac{1}{1} =1 Now, (x_{1}, y_{1}) = (a, 2 a) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 a = 1 (x - a) \Rightarrow y - 2 a = x - a \Rightarrow x - y + a = 0$

Equation of normal:

$Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - 2 a = - 1 (x - a) \Rightarrow y - 2 a = - x + a \Rightarrow x + y = 3 a$

(iv)
$x = a \sec t and y = b \tan t \frac{d x}{d t} = a \sec t \tan t and \frac{d y}{d t} = b \sec^{2} t ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{b \sec^{2} t}{a \sec t \tan t} = \frac{b}{a} \cos e c t Slope of tangent, m = {(\frac{d y}{d x})}_{t = t} = \frac{b}{a} \cos e c t Now, (x_{1}, y_{1}) = (a \sec t, b \tan t) Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - b \tan t = \frac{b}{a} \cos e c t (x - a s e c t) \Rightarrow y - \frac{b \sin t}{\cos t} = \frac{b}{a \sin t} (x - \frac{a}{\cos t}) \Rightarrow \frac{y \cos t - b \sin t}{\cos t} = \frac{b}{a \sin t} (\frac{x \cos t - a}{\cos t}) \Rightarrow y \cos t - b \sin t = \frac{b}{a \sin t} (x \cos t - a) \Rightarrow a y \sin t \cos t - a b \sin^{2} t = b x \cos t - a b \Rightarrow b x \cos t - a y \sin t \cos t - a b (1 - \sin^{2} t) = 0 \Rightarrow b x \cos t - a y \sin t \cos t = a b \cos^{2} t Dividing by \cos^{2} t, b x \sec t - a y \tan t = a b$

$Equation of normal is, y - y_{1} = m (x - x_{1}) \Rightarrow y - b \tan t = \frac{- a}{b} sin t (x - a \sec t) \Rightarrow y - b \frac{\sin t}{\cos t} = \frac{- a}{b} sin t (x - \frac{a}{\cos t}) \Rightarrow \frac{y \cos t - b \sin t}{\cos t} = \frac{- a}{b} sin t (\frac{x \cos t - a}{\cos t}) \Rightarrow y \cos t - b \sin t = \frac{- a}{b} s in t (x \cos t - a) \Rightarrow b y \cos t - b^{2} \sin t = - a x \sin t \cos t + a^{2} \sin t \Rightarrow a x \sin t \cos t + b y \cos t = (a^{2} + b^{2}) \sin t Dividing both sides by sin t, a x \cos t + b y \cot t = a^{2} + b^{2}$

(v)
$x = a (θ + \sin θ) and y = a (1 - \cos θ) \frac{d x}{d θ} = a (1 + \cos θ) and \frac{d y}{d θ} = a \sin θ ∴ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a \sin θ}{a (1 + \cos θ)} = \frac{\sin θ}{(1 + \cos θ)} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}} = \tan \frac{θ}{2} . . . (1) Slope of tangent, m = {(\frac{d y}{d x})}_{θ} = \tan \frac{θ}{2} Now, (x_{1}, y_{1}) = [a (θ + \sin θ), a (1 - \cos θ)] Equation of tangent is, y - y_{1} = m (x - x_{1}) \Rightarrow y - a (1 - \cos θ) = \tan \frac{θ}{2} [x - a (θ + \sin θ)] \Rightarrow y - a (2 \sin^{2} \frac{θ}{2}) = x \tan \frac{θ}{2} - a θ \tan \frac{θ}{2} - a \tan \frac{θ}{2} \sin θ \Rightarrow y - a (2 \sin^{2} \frac{θ}{2}) = x \tan \frac{θ}{2} - a θ \tan \frac{θ}{2} - a \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}} 2 \sin \frac{θ}{2} \cos \frac{θ}{2} (From (1)) \Rightarrow y - 2 a \sin^{2} \frac{θ}{2} = (x - a θ) \tan \frac{θ}{2} - 2 a \sin^{2} \frac{θ}{2} \Rightarrow y = (x - a θ) \tan \frac{θ}{2}$

$Equation of normal is, y - a (1 - \cos θ) = - \cot \frac{θ}{2} [x - a (θ + \sin θ)] \Rightarrow \tan \frac{θ}{2} [y - a (2 \sin^{2} \frac{θ}{2})] = - x + a θ + a \sin θ \Rightarrow \tan \frac{θ}{2} [y - a \{2 (1 - \cos^{2} \frac{θ}{2})\}] = - x + a θ + a \sin θ \Rightarrow \tan \frac{θ}{2} (y - 2 a) + a (2 \sin \frac{θ}{2} \cos \frac{θ}{2}) = - x + a θ + asin θ \Rightarrow \tan \frac{θ}{2} (y - 2 a) + a \sin θ = - x + a θ + asin θ \Rightarrow \tan \frac{θ}{2} (y - 2 a) = - x + a θ \Rightarrow \tan \frac{θ}{2} (y - 2 a) + x - a θ = 0$

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Q. Find the equations of the tangent and the normal to the following curves at the indicated points.

(i) x = θ + sinθ, y = 1 + cosθ at θ =

\frac{π}{2}

x = \frac{2 a t^{2}}{1 + t^{2}}, y = \frac{2 a t^{3}}{1 + t^{2}} at t = \frac{1}{2}

(iii) x = at², y = 2at at t = 1
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