Question

Find dy/d's if x= 15 ( t + sin t ) ; y = 18 ( 1- cos t)

Answer 1

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Q1) Find dy/dx if x=15(t+sint) and y=18(1-cost) .

$\: \: \underline{ \boxed { \bf{ \divideontimes \: answer}}}$

$\: \\ \mapsto \displaystyle \tt{x = 15(t + sint)....given \: equation}$

$\: \bullet \underline{ \bf{differentiating \: the \: equation \: with \: respect \: to \: t}}$

$\: \: \implies \displaystyle \tt{ \frac{dx}{dt} = 15t + 15sint}$

$\: \: \bullet \underline{ \bf{by \: using \: chain \: rule}}$

$\: \implies \displaystyle \tt{ \frac{dx}{dt} = 15 + 15cost}$

$\: \: \\ \implies \displaystyle \tt{ \frac{dx}{dt} = 15(1 + cost}).....(1)$

$\: \: \\ \divideontimes \boxed{ \displaystyle \tt{ \frac{dx}{dt} = 15(1 + cost)}}....(1)$

$\: \: \bullet \underline{\bf{differentiating \: the \: equation \: with \: respect \: to \: t}}$

$\: \: \\ \mapsto \displaystyle \tt{y = 18(1 - cost)}....given \: equation$

$\: \: \implies \displaystyle \tt{ \frac{dy}{dt} = 18 - 18cost}$

$\: \: \implies \displaystyle \tt{ \frac{dy}{dt} = 0 - 18 \times - sint}$

$\: \: \implies \displaystyle \tt{ \frac{dy}{dt} = 18sint}$

$\: \divideontimes \boxed{ \displaystyle \tt{ \frac{dy}{dt} = 18sint}}........(2)$

$\: \ \bullet \underline{\displaystyle \sf{dividing \: equation \: (1) \: and \: (2)}}$

$\: \: \implies \displaystyle \tt{ \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \frac{18sint}{15(1 + cost)} }$

$\: \implies \displaystyle \tt{ \frac{dy}{dx} = \cancel \frac{18}{15} \frac{sint}{(1 + cost)} }$

$\: \: \implies \displaystyle \tt{ \frac{dy}{dx} = \frac{6sint}{5 {(cos \frac{t}{2}) }^{2} }}$

$\: \\ \: \implies \boxed{ \displaystyle \tt{ \frac{dy}{dx} = \frac{6sint}{5({cos \frac{t}{2} })^{2} } }}is \: the \: answer$