Solveeit Logo
Login

Question

Question: If $x = 2 \cos \theta - \cos 2\theta$ and $y = 2 \sin \theta - \sin 2\theta$, then $\frac{d^2y}{dx^2...

If x=2cos⁡θ−cos⁡2θx = 2 \cos \theta - \cos 2\theta and y=2sin⁡θ−sin⁡2θy = 2 \sin \theta - \sin 2\theta, then d2ydx2\frac{d^2y}{dx^2} is equal to

A

32tan⁡3θ2\frac{3}{2} \tan \frac{3\theta}{2}

B

32sec⁡3θ2tan⁡3θ2\frac{3}{2} \sec \frac{3\theta}{2} \tan \frac{3\theta}{2}

C

32sec⁡23θ2\frac{3}{2} \sec^2 \frac{3\theta}{2}

D

sec⁡23θ2\sec^2 \frac{3\theta}{2}

Answer

32sec⁡23θ2\frac{3}{2} \sec^2 \frac{3\theta}{2}

Explanation

Solution

We shall show that by writing the given equations in “sum‐to‐product” form one eventually obtains

dydx=tan⁡3θ2.\frac{dy}{dx}=\tan\frac{3\theta}{2}.

Differentiating with respect to θ (using the chain‐rule!) gives

ddθ(tan⁡3θ2)=32sec⁡23θ2 .\frac{d}{d\theta}\Bigl(\tan\frac{3\theta}{2}\Bigr)=\frac{3}{2}\sec^2\frac{3\theta}{2}\,.

Then using the standard formula

d2ydx2=ddθ(dydx)dx/dθ ,\frac{d^2y}{dx^2}=\frac{\frac{d}{d\theta}\Bigl({dy\over dx}\Bigr)}{dx/d\theta}\,,

one shows (after a little algebra using the identities below) that this simplifies to

d2ydx2=32sec⁡23θ2 .\frac{d^2y}{dx^2}=\frac{3}{2}\sec^2\frac{3\theta}{2}\,.

A short outline of the solution is as follows:

:::mermaid flowchart TD A[Start with x=2cos⁡θ−cos⁡2θ,y=2sin⁡θ−sin⁡2θx=2\cos\theta-\cos2\theta,\quad y=2\sin\theta-\sin2\theta] B[Differentiate: dx/dθ=−2sin⁡θ+2sin⁡2θdx/d\theta=-2\sin\theta+2\sin2\theta, dy/dθ=2cos⁡θ−2cos⁡2θdy/d\theta=2\cos\theta-2\cos2\theta] C[Write numerator and denominator in sum‐to‐product form:] D[cos⁡θ−cos⁡2θ=2sin⁡3θ2sin⁡θ2 \cos\theta-\cos2\theta=2\sin\frac{3\theta}{2}\sin\frac{\theta}{2}] E[sin⁡2θ−sin⁡θ=2cos⁡3θ2sin⁡θ2\sin2\theta-\sin\theta=2\cos\frac{3\theta}{2}\sin\frac{\theta}{2}] F[Thus, dydx=2sin⁡3θ2sin⁡θ22cos⁡3θ2sin⁡θ2=tan⁡3θ2\frac{dy}{dx}=\frac{2\sin\frac{3\theta}{2}\sin\frac{\theta}{2}}{2\cos\frac{3\theta}{2}\sin\frac{\theta}{2}}=\tan\frac{3\theta}{2}] G[Differentiate: ddθtan⁡3θ2=32sec⁡23θ2\frac{d}{d\theta}\tan\frac{3\theta}{2}=\frac{3}{2}\sec^2\frac{3\theta}{2}] H[Divide by dx/dθdx/d\theta (which, after a similar rewriting, cancels the extra factors) to get:] I[d2ydx2=32sec⁡23θ2\frac{d^2y}{dx^2}=\frac{3}{2}\sec^2\frac{3\theta}{2}]

A-->B-->C
C-->D
C-->E
D-->F
E-->F
F-->G
G-->H
H-->I

:::

Thus the answer is

32sec⁡23θ2\boxed{\frac{3}{2}\sec^2\frac{3\theta}{2}}

which corresponds to option (C).

Explanation (minimal):

  1. Write xx and yy in forms that allow the use of sum‐to‐product identities: cos⁡θ−cos⁡2θ=2sin⁡3θ2sin⁡θ2\cos\theta-\cos2\theta=2\sin\frac{3\theta}{2}\sin\frac{\theta}{2} and sin⁡2θ−sin⁡θ=2cos⁡3θ2sin⁡θ2\sin2\theta-\sin\theta=2\cos\frac{3\theta}{2}\sin\frac{\theta}{2}.

  2. Hence, dydx=2sin⁡3θ2sin⁡θ22cos⁡3θ2sin⁡θ2=tan⁡3θ2\displaystyle \frac{dy}{dx}=\frac{2\sin\frac{3\theta}{2}\sin\frac{\theta}{2}}{2\cos\frac{3\theta}{2}\sin\frac{\theta}{2}}=\tan\frac{3\theta}{2}.

  3. Differentiating, ddθtan⁡3θ2=32sec⁡23θ2\displaystyle \frac{d}{d\theta}\tan\frac{3\theta}{2}=\frac{3}{2}\sec^2\frac{3\theta}{2}.

  4. Then using d2ydx2=d/dθ(dy/dx)dx/dθ\frac{d^2y}{dx^2}=\frac{d/d\theta(dy/dx)}{dx/d\theta} (and the similar rewriting of dx/dθdx/d\theta) one obtains d2ydx2=32sec⁡23θ2\displaystyle \frac{d^2y}{dx^2}=\frac{3}{2}\sec^2\frac{3\theta}{2}.