Ok one more parametric equation problem.

[shivers20]

x= 2cost, y= 2sint, t = pi/4

Find 2nd derivatives.

Here goes,

formula:
[(d/dt) (dy/dx)] / (dx/dt)=

(dx/dt) (d^2y/dt^2) - (dy) (d^2x)/(dt) (dt^2) / (dx/dt^3)

Now lets see if i can substitute the formula with my numbers.

d/dt = 0, since my slope was -1 from the first derivative?
multiply the 0 and dy/dt=2cost

divide by dx/dt=-2sint

equals -2sint then multiply d^2y/dt^2= -sin I am already lost haha! Can anyone please give me a hand with this problem, I am not looking for the answer, I would be happy if I could find an example on the internet with using cos or sin. I only found problems that deal with numbers and it isnt very clear. My overpriced book has one example for each.

 

[shivers20]

Ok so I have been doing some more research and I came up with this.

2cos(t) / -2 sin(t) now i use the quotient rule: (2 cost)(-2cost) - (-2sint)(-2sint)

equals -4cos^2x- 4sin^2x

how do i add these up. cos^2x-sin^2x= 1?

Am I on the right track?

 

[skeeter]

\(\displaystyle \L y(t) = 2sint\)
\(\displaystyle \L x(t) = 2cost\)

\(\displaystyle \L y'(t) = 2cost\)
\(\displaystyle \L x'(t) = -2sint\)

\(\displaystyle \L y''(t) = -2sint\)
\(\displaystyle \L x''(t) = -2cost\)

if \(\displaystyle \L \frac{dy}{dx} = \frac{y'(t)}{x'(t)}\), then

\(\displaystyle \L \frac{d^2y}{dx^2} = \frac{x'(t) y''(t) - y'(t) x''(t)}{[x'(t)]^3}\)


\(\displaystyle \L \frac{d^2y}{dx^2} = \frac{4sin^2t + 4cos^2t}{-8sin^3t} = \frac{1}{-2sin^3t}\)

now sub in \(\displaystyle \L \frac{\pi}{4}\) for t.

 

[soroban]

Hello, shivers20!

\(\displaystyle \L\begin{array}{cc}x\:=\:2\cos\theta \\ y\:= \:2\sin\theta\end{array}\;\;\theta\,=\,\frac{\pi}{4}\)

Find 2nd derivative.

. . . . . . . . . Formulas

\(\displaystyle \L\:\frac{dy}{dx}\:=\:\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\) . . . . . \(\displaystyle \L\:\frac{d^2y}{dx^2}\:=\:\frac{\frac{d}{d\theta}\left(\frac{dy}{dx}\right)}{\frac{dx}{d\theta}}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: \(\displaystyle \L\:\begin{array}{cc}x\:=\:2\cos\theta\;&\;\Rightarrow\;&\;\frac{dx}{dt}\:=\:-2\sin\theta \\ y\:=\:2\sin\theta\;&\;\Rightarrow\;&\frac{dy}{dt}\:=\:2\cos\theta\end{array}\)


Hence: \(\displaystyle \L\:\frac{dy}{dx} \:=\:\frac{2\cos\theta}{-2\sin\theta} \:=\:-\cot\theta\)

And: \(\displaystyle \L\:\frac{d^2y}{dx^2}\:=\:\frac{\frac{d}{d\theta}(-\cot\theta)}{\frac{dx}{d\theta}} \:=\:\frac{\csc^2\theta}{-2\sin\theta} \:=\:-\frac{1}{2}\csc^3\theta\)


At \(\displaystyle \theta\,=\,\frac{\pi}{4}\)

. . \(\displaystyle \L\frac{dy}{dx}\:=\:-\cot\left(\frac{\pi}{4}\right)\:=\:-1\)

. . \(\displaystyle \L\frac{d^2y}{dx^2}\:=\:-\frac{1}{2}\csc^3\left(\frac{\pi}{4}\right)\:=\:-\frac{1}{2}\left(\sqrt{2}\right)^3\:=\:-\sqrt{2}\)