Find $\frac{d^2y}{dx^2}$ for Parametric Equations x,y

[ILoveBaseball]

find \frac{d^2y}{dx^2} as a function of t, for the given the parametric equations:

x = 2-4*cos(t)
y= 4+cos(t)^2

\frac{d^2y}{dx^2} = _______

dy/dt = -2*cos(t)*sin(t)
second derv. 2*sin(t)^2-2*cos(t)^2

dx/dt = 4*sin(t)
second derv. 4*cos(t)

\frac{d^2y}{dx^2} = \frac{2*sin(t)^2-2*cos(t)^2}{4*cos(t)}

what did i do wrong?

 

[Oxymoron]

\frac{d^2y}{dx^2} \neq \frac{\frac{d^2y}{dt^2}}{\frac{d^2x}{dt^2}}

Here is how I would do it.

x = 2 - 4\cos(t)
4\cos(t) = 2-x
t = \arccos\left(\frac{2-x}{4}\right)

Now we have t explicitly.

Substituite this back in for t and we have

y = 4 + \cos\left(\arccos\left(\frac{2-x}{4}\right)\right)^2
\sqrt{y} = 2 + \cos\left(\arccos\left(\frac{2-x}{4}\right)\right)
\sqrt{y} = 2 + \left(\frac{2-x}{4}\right)
\sqrt{y} = \frac{3}{2} - \frac{x}{4}
y = \frac{9}{4} - \frac{x^2}{16}

Now differentiate.

\frac{dy}{dx} = -\frac{x}{8}
\frac{d^2y}{dx^2} = -\frac{1}{8}

 

[whozum]

Also in general, for y(t) and x(t),

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

\frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{(\frac{dx}{dt})^2}

Simplifying the RHS of the second equation you get:

\frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{(\frac{dx}{dt})^2} = \frac{d^2y}{dt^2} \frac{dt^2}{dx^2} and by cancelling you get the desired result \frac{d^2y}{dx^2}

 

[ILoveBaseball]

awesome thanks

 

[xanthym]

Also in general, for y(t) and x(t),

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \ \ \ \color{blue} \longleftarrow \mathbf{ (EQ \ 1) }

\color{red} \frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{(\frac{dx}{dt})^2} \ \ \ \mathbf{ \longleftarrow ( \underline{NOT\ \ CORRECT})}

Simplifying the RHS of the second equation you get:

\frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{(\frac{dx}{dt})^2} = \frac{d^2y}{dt^2} \frac{dt^2}{dx^2} and by cancelling you get the desired result \frac{d^2y}{dx^2}

whozum:
Highlighted equation above for (d²y/dx²) is NOT correct. For parametric equations {y=y(t) & x=x(t)}, the second derivative is given by:

:(2): \ \ \ \ \ \color{blue} \frac{d^2y}{dx^2} \ = \ \left( \frac{d}{dt} \left( \frac{dy}{dx} \right) \right) / \left( \frac{dx}{dt} \right)

where (dy/dx) is function of "t" obtained from the correct Equation #1 above.


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