Question

X= tcost , y= t+sint find dx/dy at t=pi/2

Answer 1

Given:

x= tcost

y= t+sint

To Find:

Value of $\rm{\dfrac{dx}{dy}}$ at $\rm{t=\dfrac{\pi}{2}}$

Solution:

Firstly, we have to differentiate x with respect to t

On differentiating x w.r.t. t, we get

$\longrightarrow\rm{\dfrac{dx}{dt}=t(-sint)+cost(1)}$

$\longrightarrow\rm{\dfrac{dx}{dt}=-tsint+cost}-----(1)$

Now, we have to differentiate y with respect to t

On differentiating y w.r.t. t, we get

$\longrightarrow\rm{\dfrac{dy}{dt}=1+cost}-----(2)$

On dividing (1) by (2), we get

$\longrightarrow\rm{\dfrac{\dfrac{dx}{\cancel{dt}}}{\dfrac{dy}{\cancel{dt}}}=\dfrac{-tsint+cost}{1+cost} }$

$\longrightarrow\rm{\dfrac{dx}{dy}=\dfrac{-tsint+cost}{1+cost}}$

Now, on putting value of t in above equation, we get

$\longrightarrow\rm{\bigg(\dfrac{dx}{dy}\bigg)_{t=\frac{\pi}{2}}=\dfrac{-\dfrac{\pi}{2}(sin\dfrac{\pi}{2})+cos\dfrac{\pi}{2}}{1+cos\dfrac{\pi}{2}}}$

$\longrightarrow\rm{\bigg(\dfrac{dx}{dy}\bigg)_{t=\frac{\pi}{2}}=\dfrac{-\dfrac{\pi}{2}(1)+0}{1+0}}$

$\longrightarrow\rm{\bigg(\dfrac{dx}{dy}\bigg)_{t=\frac{\pi}{2}}=\dfrac{-\dfrac{\pi}{2}}{1}}$

$\longrightarrow\rm\green{\bigg(\dfrac{dx}{dy}\bigg)_{t=\frac{\pi}{2}}=-\dfrac{\pi}{2}}$

Formulae to Remember

$\longrightarrow\rm{\dfrac{d}{dx}(x^{n})=nx^{n-1}}$

$\longrightarrow\rm{\dfrac{d}{dx}(sinx)=cosx}$

$\longrightarrow\rm{\dfrac{d}{dx}(cosx)=-sinx}$

$\longrightarrow\rm{\dfrac{d}{dx}(tanx)=sec^{2}x}$

$\longrightarrow\rm{\dfrac{d}{dx}(cotx)=-cosec^{2}x}$

$\longrightarrow\rm{\dfrac{d}{dx}(secx)=secx.tanx}$

$\longrightarrow\rm{\dfrac{d}{dx}(cosecx)=-cosecx.cotx}$

$\longrightarrow\rm{\dfrac{d}{dx}(ln\:x)=\dfrac{1}{x}}$

$\longrightarrow\rm{\dfrac{d}{dx}(e^{x})=e^{x}}$

$\longrightarrow\rm{\dfrac{d}{dx}(f(x))=f'(x)}$

Answer 2

GIVEN :-

$\sf{x = t . cos(t)}$

$\sf{y = t + sin(t)}$

$\sf{t=\dfrac{\pi}{2} }$

TO EVALUATE :-

$\sf{\dfrac{dx}{dy} }$

SOLUTION :-

$\sf{\dfrac{dx}{dt}=\dfrac{d}{dt}[tcos(t)]}$

$\sf{\implies \dfrac{dx}{dt}=\dfrac{d}{dt}(t).cos(t)+t.\dfrac{d}{dt}[cos(t)] }$

$\displaystyle{\sf{\implies \frac{dx}{dt}=1.cos(t)+t[-sin(t)] }}\\\\\displaystyle{\sf{\implies \frac{dx}{dy}=cos(t)-tsin(t) }}$

$\rule{150}2$

$\displaystyle{\sf{\frac{dy}{dt}=\frac{d}{dt}[t+sin(t)] }}$

$\displaystyle{\sf{\implies \frac{dy}{dt}=\frac{d}{dt}(t)+\frac{d}{dt}[sin(t)] }}\\\\\displaystyle{\sf{\implies \frac{dy}{dt}=1+cos(t) }}$

$\rule{150}2$

$\displaystyle{\sf{Now,}}$

$\displaystyle{\sf{\frac{dx}{dy} =\frac{dx}{dt}\div \frac{dy}{dt} }}\\\\\boxed{\displaystyle{\sf{\implies \frac{dx}{dy}=\frac{cos(t)-tsin(t)}{1+cos(t)} }}}$

$\textsf{Now putting the value of t as }\sf{\dfrac{\pi}{2} :- }$

$\displaystyle{\sf{\implies \frac{dx}{dy}=\frac{cos(\frac{\pi}{2} )-tsin(\frac{\pi}{2} )}{1+cos(\frac{\pi}{2} )} }}$

$\displaystyle{\sf{\implies \frac{dx}{dy}=\frac{0 -t \times 1}{1+0} }}$

$\displaystyle{\sf{\implies \frac{dx}{dy}=\frac{-t}{1} }}$

$\underline{\underline{\boxed{\boxed{\displaystyle{\sf{\implies \frac{dx}{dy}=-t=\frac{-\pi}{2} }}}}}}$

$\rule{150}2$

$\underline{\sf{Formulas\ applied\ :-}}$

$\bullet\ \textsf{Product rule :-}\sf{[a(x)+b(y)]'=a'(x).b(x)+a(y)+b'(y)}$

$\bullet\ \textsf{Differentiation of a variable is 1.}$

$\bullet\ \textsf{Differentiation of sin(x) is cos(x).}$

$\bullet\ \textsf{Differentiation of cos(x) is -sin(x).}$