How to find the second derivative of a parametric curve
Formula for the second derivative of a parametric curve
To find the second derivative of a parametric curve, we need to find its first derivative ???dy/dx???, using the formula
???\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}???
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and then plug it into this formula for the second derivative:
???\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}???
where ???d^2y/dx^2??? is the second derivative of the parametric curve, ???dy/dx??? is its first derivative and ???dx/dt??? is the first derivative of the equation for ???x???. The ???d/dt??? is notation that tells us to take the derivative of ???dy/dx??? with respect to ???t???.
How to calculate the second derivative of a parametric curve
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Calculating the second derivative of the parametric equation
Example
Find the second derivative of the parametric curve.
???x=5t^3+6t???
???y=t^4-3???
To find the first derivative, we’ll solve for ???dx/dt??? and ???dy/dt???.
This means we will have to solve for ???dy/dt???, ???dx/dt??? and ???dy/dx??? first. Let’s start with ???dy/dt???.
???x=5t^3+6t???
???\frac{dx}{dt}=15t^2+6???
and
???y=t^4-3???
???\frac{dy}{dt}=4t^3???
Plugging these two derivatives into the formula for the first derivative,
???\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}???
we get
???\frac{dy}{dx}=\frac{4t^3}{15t^2+6}???
The d/dt is notation that tells us to take the derivative of dy/dx with respect to t.
Now plugging the first derivative ???dy/dx??? and the value we found earlier for ???dx/dt??? into the formula for the second derivative,
???\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}???
we get
???\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{4t^3}{15t^2+6}\right)}{15t^2+6}???
We’ll use quotient rule to take the derivative of ???dy/dx??? with respect to ???t???.
???\frac{d^2y}{dx^2}=\frac{\frac{\left(12t^2\right)\left(15t^2+6\right)-\left(4t^3\right)\left(30t\right)}{\left(15t^2+6\right)^2}}{15t^2+6}???
???\frac{d^2y}{dx^2}=\frac{\left(12t^2\right)\left(15t^2+6\right)-\left(4t^3\right)\left(30t\right)}{\left(15t^2+6\right)^2}\cdot\frac{1}{15t^2+6}???
???\frac{d^2y}{dx^2}=\frac{180t^4+72t^2-120t^4}{\left(15t^2+6\right)^3}???
???\frac{d^2y}{dx^2}=\frac{60t^4+72t^2}{\left(15t^2+6\right)^3}???
???\frac{d^2y}{dx^2}=\frac{12t^2\left(5t^2+6\right)}{\left(15t^2+6\right)^3}???
