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)
I don't understand this notation, please help!
Hello, Del!
You don't understand $\displaystyle \frac{d^2y}{dx^2}\:?\quad\hdots$ It's the second derivative of $\displaystyle y.$Find $\displaystyle \frac{d^2y}{dx^2}$, as a fuction of t.
. . $\displaystyle \begin{array}{ccc}x &= &3 - 8\cos t \\
y & = & 5 + \cos^3\!t \end{array}$
I don't understand this notation. . . . . Which notation?
You don't understand $\displaystyle \cos^3 t\:?\quad\hdots$ It's $\displaystyle (\cos t)^3$
(a) Find the first derivative.
Formula: .$\displaystyle \frac{dy}{dx} \;=\;\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} $
It says:
. . (1) Differentiate $\displaystyle y$ with respect to $\displaystyle t.$
. . (2) Differentiate $\displaystyle x$ with respect to $\displaystyle t.$
. . (3) Form the fraction.
(b) Find the second derivative.
Formula: .$\displaystyle \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 $\displaystyle \frac{dy}{dx}$ . . . which we found in part (a).
The bottom says: divide by $\displaystyle \frac{dx}{dt}$ . . . which we also found in part (a).
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Back to the problem: .$\displaystyle \begin{array}{ccc} x &=& 3-8\cos t \\ y &=& 5 + \cos^3\!t \end{array}$
(a) . $\displaystyle \frac{dy}{dt} \:=\:\text{-}3\cos^2\!t\sin t\qquad\qquad\frac{dx}{dt} \:=\:8\sin t $
. . . .$\displaystyle \frac{dy}{dx} \;=\;\frac{\text{-}3\cos^2\!t\sin t}{8\sin t} \quad\Rightarrow\quad\boxed{\frac{dy}{dx}\;=\;\tex t{-}\frac{3}{8}\cos^2\!t}$
(b) .$\displaystyle \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: . $\displaystyle \frac{d^2y}{dx^2} \;=\;\frac{\frac{3}{4}\cos t\sin t}{8\sin t} \;=\;\boxed{\frac{3}{32}\cos t} $