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EXERCISE 1.4 · Q161

Q.Find dydx\dfrac{dy}{dx} if x=t2+t+1x=t^2+t+1, y=sin⁡πt2+cos⁡πt2y=\sin\dfrac{\pi t}{2}+\cos\dfrac{\pi t}{2}, at t=1t=1.

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We have x=t2+t+1x=t^2+t+1, y=sin⁡πt2+cos⁡πt2y=\sin\dfrac{\pi t}{2}+\cos\dfrac{\pi t}{2}.

Step 1. dxdt=2t+1\dfrac{dx}{dt}=2t+1

Step 2.

dydt=π2cos⁡πt2−π2sin⁡πt2=π2(cos⁡πt2−sin⁡πt2)\frac{dy}{dt}=\frac{\pi}{2}\cos\frac{\pi t}{2}-\frac{\pi}{2}\sin\frac{\pi t}{2}=\frac{\pi}{2}\left(\cos\frac{\pi t}{2}-\sin\frac{\pi t}{2}\right)

Step 3. At t=1t=1: dxdt=2(1)+1=3\dfrac{dx}{dt}=2(1)+1=3, and cos⁡π2=0\cos\dfrac{\pi}{2}=0, sin⁡π2=1\sin\dfrac{\pi}{2}=1, so …

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