1. ## More Parametric Equations

Find d^2y/dx^2 , as a fuction of t, for the given the parametric equations:

x = 3 - 8cos(t)

y = 5 + cos^3(t)

2. Originally Posted by Del
Find d^2y/dx^2 , as a fuction of t, for the given the parametric equations:

x = 3 - 8cos(t)

y = 5 + cos^3(t)

It is the 2nd derivative of y with respect to x.

$t = arccos\left(\frac{3-x}{8}\right)$

Plug that into y to get an equation with x as the parameter and no t's.

$y = 5 + \left(\frac{3-x}{8}\right)^3$

Now, take the 2nd derivative...

3. Originally Posted by Del
Find d^2y/dx^2 , as a fuction of t, for the given the parametric equations:

x = 3 - 8cos(t)

y = 5 + cos^3(t)

-Dan

4. Hello, Del!

Find $\frac{d^2y}{dx^2}$, as a fuction of t.

. . $\begin{array}{ccc}x &= &3 - 8\cos t \\
y & = & 5 + \cos^3\!t \end{array}$

I don't understand this notation. . . . . Which notation?
You don't understand $\frac{d^2y}{dx^2}\:?\quad\hdots$ It's the second derivative of $y.$

You don't understand $\cos^3 t\:?\quad\hdots$ It's $(\cos t)^3$

(a) Find the first derivative.

Formula: . $\frac{dy}{dx} \;=\;\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$
It says:
. . (1) Differentiate $y$ with respect to $t.$
. . (2) Differentiate $x$ with respect to $t.$
. . (3) Form the fraction.

(b) Find the second derivative.

Formula: . $\frac{d^2y}{dx^2} \;=\;\frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right )}{\dfrac{dx}{dt}}$

Read this formula very carefully . . .

The numerator says: differentiate $\frac{dy}{dx}$ . . . which we found in part (a).
The bottom says: divide by $\frac{dx}{dt}$ . . . which we also found in part (a).

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

Back to the problem: . $\begin{array}{ccc} x &=& 3-8\cos t \\ y &=& 5 + \cos^3\!t \end{array}$

(a) . $\frac{dy}{dt} \:=\:\text{-}3\cos^2\!t\sin t\qquad\qquad\frac{dx}{dt} \:=\:8\sin t$

. . . . $\frac{dy}{dx} \;=\;\frac{\text{-}3\cos^2\!t\sin t}{8\sin t} \quad\Rightarrow\quad\boxed{\frac{dy}{dx}\;=\;\tex t{-}\frac{3}{8}\cos^2\!t}$

(b) . $\frac{d}{dt}\left(\frac{dy}{dx}\right) \;=\;\frac{d}{dt}\left(\text{-}\frac{3}{8}\cos^2\!t\right) \;=\;\frac{3}{4}\cos t\sin t$

Therefore: . $\frac{d^2y}{dx^2} \;=\;\frac{\frac{3}{4}\cos t\sin t}{8\sin t} \;=\;\boxed{\frac{3}{32}\cos t}$