Assume a curve is given by the parametric equations


Assume a curve is given by the parametric equations x = g(t) and y = h(t), where g and h are twice differentiable. Use the Chain Rule to show that

y^{\prime \prime}(x)=\frac{x^{\prime}(t)y^{\prime \prime}(t)-y^{\prime}(t)x^{\prime \prime}(t)}{(x^{\prime}(t))^{3}}

Solution:

 

x=g(t)

 

y=h(t)

\Rightarrow y^{\prime}(x) =\frac{dy}{dx} = \frac{h^{\prime}(t)}{g^{\prime}(t)}=\frac{\frac{dh}{dt}}{\frac{dg}{dt}}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}

 

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

 

y^{\prime \prime}(x)=\frac{\text{d}(\frac{\text{d}y}{\text{d}x})}{\text{d}x}=\frac{d(\frac{\frac{dy}{dt}}{\frac{\text{d}x}{\text{d}t}})}{dx}

 

y^{\prime\prime}(x)=\frac{\left[{\frac{d}{dx}(\frac{dy}{dt})\cdot\frac{dx}{dt}-\frac{d}{dx}(\frac{dx}{dt})\cdot\frac{dy}{dt}}{}\right]\cdot\frac{\frac{dx}{dt}}{\frac{dx}{dt}}}{(\frac{dx}{dt})^{2}} (Expand the numerator)

 

{y^{\prime\prime}(x)=\frac{\left[{\frac{d}{\color{red}{dx}}(\frac{dy}{dt})\cdot\frac{\color{red}{dx}}{dt}\cdot\frac{dx}{dt}-\frac{d}{\color{red}{dx}}(\frac{dx}{dt})\cdot\frac{dy}{dt}\cdot\frac{\color{red}{dx}}{dt}}{}\right]}{(\frac{dx}{dt})^{3}}} (Cancel out like terms)

 

{y^{\prime\prime}(x)=\frac{\left[{\frac{d}{dt}(\frac{dy}{dt})\cdot\frac{dx}{dt}-\frac{d}{dt}(\frac{dx}{dt})\cdot\frac{dy}{dt}}{}\right]}{(\frac{dx}{dt})^{3}}}

 

\therefore y^{\prime \prime}(x)=\frac{x^{\prime}(t)y^{\prime \prime}(t)-y^{\prime}(t)x^{\prime \prime}(t)}{(x^{\prime}(t))^{3}}

 

If you have a hard time following the math text, try following this text:

Assume a curve is given by the parametric equations

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