# Prove that $y''(0) = -1, x = \cos\left(\frac{t}{1+t}\right), y = \sin\left(\frac{t}{1+t}\right)$

My attempt:

$$x'= -\sin\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}$$

$$y'= \cos\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}$$

then I attempted to divide y' on x' which resulted in

$$-\cot\left(\frac{t}{1+t}\right)$$

When I try to take the double dash of this it gives me

$$\csc^2\left(\frac{t}{1+t}\right),$$

but this function is undefined at 0.

• Assuming you mean $d^2y/dx^2\Big|_{x=0}$, this derivative is undefined as the function is not univocal.
– user65203
Dec 5, 2014 at 17:41
• Good luck on proving the three equations $y''(0)=0$, $x=\cos(\frac t{1+t})$, and $y=\sin(\frac t{1+t})$ with no hypotheses.
– bof
Dec 6, 2014 at 5:15

The second derivative of a parametric curves is \begin{align*} &\frac{d}{dt} \left( \frac{dy}{dx} \right) / \left( \frac{dx}{dt} \right)\\ =&\frac{d}{dt} \left( -\cot \left( \frac{t}{1+t} \right) \right) / \left( \frac{d}{dt} \left( \sin\left( \frac{t}{1+t} \right) \right) \right)\\ =&\frac{\csc^2\left( \frac{t}{1+t} \right)}{\left( 1+t \right)^2} \cdot \frac{\left( 1+t \right)^2}{-\sin\left( \frac{t}{1+t} \right)}\\ =&-\csc^3\left( \frac{t}{1+t} \right) \end{align*} This is not defined for $t=0$.

You can see this easily by $\frac{t}{1+t}\mapsto t$. Now you can more clearly see that it traces the unit circle with $t=0$ giving the point $(0,1)$. Intuitively, the slope is infinitely great at this point.

Assuming primes are with respect to t, direct differentiation of $y'$ gives:

$-\dfrac {(2 (1 + t) \cos[t/(1 + t)] + \sin[t/(1 + t)])} {(1 + t)^4}$

which gives you a different result:

$y^{''}(0)=-2$

EDIT1:

For t > 0, the unit circle exists only on the arc $[(0.6,0.8)-(1,0)]$, approximately given on the plot. Negative arguments of $t$ unexpectedly deliver a unit circular billiards board :) ; the above plots were made on domain $t$: $[( 0 <t <12),( -2 <t <0 ) ]$. The trouble spot $t=-1$ is to be tackled with a change of variable.

If I were you, I would offer a bounty for correct domain definitions to go out of the puzzling situation, so that $y''(0)$ can be defined. At this moment it appears that it cannot be defined (by the participants so far).

$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$$

I think I have found the solution

$$x'= -\sin\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}$$

$$y'= \cos\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}$$

$$y''= \frac{\frac{dy}{dx}(\cos\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2})}{-\sin\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}}$$

$$y''= \frac{-\sin\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}*\frac{1}{(1+t)^2}+-2(1+t)(\cos{\frac{t}{1+t}}*\frac{1}{(1+t)^2})}{-\sin\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}}$$

When I calculate this as it is it still has 0 on the bottom but when I cancel out the -sin up and down it gives me -1, is my solution correct?

Edit: I can't cancel it out because of the + I overlooked it, I still don't know the solution

Edit: I did another mistake, I should differentiate of dy/dx(dy/dt/dx/dt) / dx/dt not just dy/dt

I emailed my professor and he said that the question is incorrect at it is in fact undefined. I am very sorry

To prove y''(0)=-1

Take the derivative of the second equation using the product, quotient and chain rules. Then replace $t$ with $0$ and evaluate.

For the other two you can integrate:

$\int -\sin(\frac{t}{1+t})*\frac{1}{(1+t)^2}*dt$

$u=\frac{t}{1+t}$

$du=\frac{1}{(1+t)^2}*dt$

$\int -\sin(u)*du$

Which I think you can evaluate :)