Question

यदि x= a(cos t + t sin t) और y=a(sin t -t cos t), तो d^{2}y/dx^{2} ज्ञात कीजिए।

Answer 1

$\textbf{Given:}$

$x=a(cos\;t+t\;sin\;t)$

$y=a(sin\;t-t\;cos\;t)$

$\frac{dx}{dt}=a[-sin\;t+t(cos\;t)+(sin\;t)1]$

$\frac{dx}{dt}=a[-sin\;t+t\;cos\;t+sin\;t]$

$\frac{dx}{dt}=at\;cos\;t$

$\text{and}$

$\frac{dy}{dt}=a[cos\;t-t(-sin\;t)-(cos\;t)1]$

$\frac{dy}{dt}=a[cos\;t+t\;sin\;t-cos\;t]$

$\frac{dy}{dt}=at\;sin\;t$

$\text{Now}$

$\displaystyle\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$\displaystyle\frac{dy}{dx}=\frac{at\;sin\;t}{at\;cos\;t}$

$\displaystyle\frac{dy}{dx}=tan\;t$

$\text{Differentiate with respect to x, we get}$

$\displaystyle\frac{d^2y}{dx^2}=\frac{d(tan\;t)}{dx}$

$\displaystyle\frac{d^2y}{dx^2}=sec^2t\;\frac{dt}{dx}$

$\displaystyle\frac{d^2y}{dx^2}=sec^2t(\frac{1}{at\;cos\;t})$

$\implies\boxed{\bf\frac{d^2y}{dx^2}=\frac{1}{at}sec^3t}$

Find more:

Find d^2y / dx^2 if. 1) x=3 cost - 2 cos^3 t & y= 3sint - 2 Sin^3t

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Answer 2

d²y/dx²  = Sec³t /at यदि x= a(cos t + t sin t) और y=a(sin t -t cos t),

Step-by-step explanation:

x= a(cos t + t sin t)

y = a(sin t -t cos t)

x= a(cos t + t sin t)

dx/dt  = a( - Sint  + tCost + Sint)

=> dx/dt = atCost

y = a(sin t -t cos t)

=> dy/dt = a(Cost - t(-sint) - Cost)

=> dy/dt = atSint

(dy/dt)/(dx/dt)  = atSint/(atCost)

=>  dy/dx =   Tant

d²y/dx²  = Sec²t (dt/dx)

=> d²y/dx²  = Sec²t /(dx/dt)

=>  d²y/dx²  = Sec²t /atCost

=> d²y/dx²  = Sec³t /at

और अधिक जानें :

(x + 3)^{2} .(x + 4)^{3} .(x + 5)^{4} प्रदत्त फलनों का x के सापेक्ष अवकलन कीजिए

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f(x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) द्वारा प्रदत्त फलन का अवकलज ज्ञात कीजिए और इस प्रकार f'(1) ज्ञात कीजिए।

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