Skip to content
EXERCISE 1.4 · Q162

Q.Find dydx\dfrac{dy}{dx} if x=2cos⁡t+cos⁡2tx=2\cos t+\cos 2t, y=2sin⁡t−sin⁡2ty=2\sin t-\sin 2t, at t=π4t=\dfrac{\pi}{4}.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
55% · 162/293 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

We have x=2cos⁡t+cos⁡2tx=2\cos t+\cos2t, y=2sin⁡t−sin⁡2ty=2\sin t-\sin2t.

Step 1. dxdt=−2sin⁡t−2sin⁡2t\dfrac{dx}{dt}=-2\sin t-2\sin2t

Step 2. dydt=2cos⁡t−2cos⁡2t\dfrac{dy}{dt}=2\cos t-2\cos2t

Step 3. At t=π4t=\dfrac{\pi}{4}: sin⁡π4=22\sin\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}, sin⁡π2=1\sin\dfrac{\pi}{2}=1, cos⁡π4=22\cos\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}, cos⁡π2=0\cos\dfrac{\pi}{2}=0.

dxdt=−2⋅22−2(1)=−2−2\frac{dx}{dt}=-2\cdot\frac{\sqrt2}{2}-2(1)=-\sqrt2-2

dydt=2⋅22−2(0)=2\frac{dy}{dt}=2\cdot\frac{\sqrt2}{2}-2(0)=\sqrt2

Step 4.

dydx=2−2−2=−22+2\frac{dy}{dx}=\frac{\sqrt2}{-\sqrt2-2}=\frac{-\sqrt2}{\sqrt2+2} …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.