Question

Find d^2y/dx^2 when x=acost and y=bsint​

Answer 1

EXPLANATION.

⇒ x = a cos(t)  &  y = b sin(t).

As we know that,

⇒ x = a cos(t).

Differentiate w.r.t t, we get.

⇒ dx/dt = a.(-sin (t)).

⇒ dx/dt = - a sin(t).

⇒ y = b sin(t).

Differentiate w.r.t t, we get.

⇒ dy/dt = b cos(t).

⇒ dy/dx = dy/dt X dx/dt.

⇒ dy/dx = b cos(t)/-a sin(t).

⇒ dy/dx = -b cot(t)/a.

Again differentiate w.r.t x, we get.

⇒ d²y/dx² = -b/a X (-cosec²t) dt/dx.

As we know that,

⇒ dx/dt = - a sin(t).

⇒ dt/dx = 1/- a sin(t).

⇒ d²y/dx² = -b/a X (-cosec²t).(1/- a sin(t)).

⇒ d²y/dx² = -b(cosec³t)/a².

                                                                                                                     

MORE INFORMATION.

Differential coefficient of some standard function.

(1) = d(constant)/dx = 0.

(2) = d(ax)/dx = a.

(3) = d(xⁿ)dx = nxⁿ⁻¹.

(4) = d(eˣ)dx = eˣ.

(5) = d(aˣ)/dx = aˣ㏒(a).

(6) = d(㏒(x))/dx = 1/x.

(7) = d(㏒ₐx)/dx = 1/x㏒(a).

Answer 2

★ Given:-

• ☯ x = a cos t
• ☯ y = b sin t

★ To Find:-

• ☯ $\sf \dfrac{d \: {}^{2} \: y}{dx \: {}^{2} }$

★ Solution:-

☯ First find $\sf \dfrac{dy}{dx}$ and if we find $\sf \dfrac{dy}{dx}$ we also

have to find $\sf \dfrac{dx}{dt}$ and $\sf \dfrac{dy}{dt}$.

☯ Finding dx/dt,

☯ x = a cos t

➙ dx/dt = -a sin t.

☯ Finding dy/dt,

☯ y = b sin t

➙ dy/dt = b cos t.

☯ Finding dy/dx ,

$\implies \tt \dfrac{dy}{dx} = \: \dfrac{dy}{dt} \times \dfrac{dt}{dx} \: = \dfrac{b \: cos \: t}{ - a \: sin \: t} \\\\ \implies\tt\dfrac{dy}{dt} = \dfrac{ - b}{a} cot \: t \\\\ \tt \implies \dfrac{d {}^{2} y}{dx {}^{2} } = \frac{ - b}{a} ( - cosec {}^{2} t). \frac{dt}{dx} \\\\ \implies\tt\frac{b}{a} (cosec {}^{2} t).( \frac{1}{a \: sin \: t} ) \\\\ \implies \tt\frac{d {}^{2}y }{dx {}^{2} } = \frac{ - b \: cosec {}^{3}t }{a { }^{2} } .$

Answer:-

$\bf\dfrac{d {}^{2}y }{dx {}^{2} } = \dfrac{ - b \: cosec {}^{3}t }{a { }^{2} } .$

★ Hope it helps u ★