Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given that $x=1+\sin(t)$ , $y=\sin(t) -\frac{1}{2} \cos(2t)$ show that $\frac{\text{d}^2y}{\text{d}x^2}=2$. I am having trouble proving this. Here is my working so far:

\begin{align}\frac{dx}{dt}&= cos(t)\\ \frac{dy}{dt}&= cos(t) + sin(2t)\end{align}

\begin{align}\frac{dy}{dx}&=\frac{\cos(t) + sin(2t)}{cos(t)}\\ \frac{d^2y}{dx^2}&=\frac{2cos(2t)cos(t) - sin(2t)sin(t)}{cos^2(t)}\frac{1}{cos(t)} \end{align}

I think its just a matter of simplifying my expression for $\frac{\text{d}^2y}{\text{d}x^2}$ using trigonometric identities but I can't see the right ones to use.

share|improve this question
    
@RGB: read the OP's work below; the edit should have been clear from that. –  Ron Gordon Aug 5 '13 at 13:48

2 Answers 2

up vote 2 down vote accepted

$$\frac{dx}{dt}=\cos t, \frac{dy}{dt}=\cos t+\sin2t=\cos t(1+2\sin t) $$

Using Chain Rule (1,2),

$$\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=1+2\sin t (\text{ assuming }\cos t\ne0) $$

Again using Chain Rule,

$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)/\frac{dx}{dt}=\frac{d(1+2\sin t)}{dt}/ \cos t=\frac{2\cos t}{\cos t}=2$$

share|improve this answer
    
Thanks for this! I am pretty weak at spotting when certain trigonometric relationships are useful, I need more practice I guess! –  ScottMillsFan Aug 5 '13 at 13:58
    
@ScottMillsFan, My pleasure. Trigonometric identities are indispensable as well as useful calculus. –  lab bhattacharjee Aug 5 '13 at 13:59

Use $\cos{2 t} = 1 - 2 \sin^2{t}$ and $\sin{t}=x-1$. Then

$$y = (x-1)^2+x-1-\frac12 = x^2-x-\frac12$$

It should be clear what to do from here.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.