# Higher Order Derivatives of Parametric Equations

Here you will learn higher order derivatives of parametric equations with examples.

Let’s begin –

## Higher Order Derivatives of Parametric Equations

We know that the differentiation of parametric equations of type x = at and y = 2at is given by formula

$$dy\over dx$$ = $$dy/dt\over dx/dt$$

where t is the parameter

Now, by again differentiating the above equation we obtain 2nd order derivative given below:

$$d^2y\over dx^2$$ = $$d\over dx$$($$dy\over dx$$)

Example : find $$d^2y\over dx^2$$ , if x = $$at^2$$ , y = 2at.

Solution : We have,

x = $$at^2$$ , y = 2at.

Differentiating both sides with respect to t,

$$\implies$$ $$dx\over dt$$ = 2at and $$dy\over dt$$ = 2a           …….(i)

$$\therefore$$ $$dy\over dx$$ = $$dy/dt\over dx/dt$$ = $$2a\over 2at$$ = $$1\over t$$

Differentiating both sides with respect to x, we get

$$d^2y\over dx^2$$ = $$d\over dx$$($$1\over t$$)

$$\implies$$ $$d^2y\over dx^2$$ = $$-1\over t^2$$$$dt\over dx$$

from (i), [$$dx\over dt$$ = 2at $$\therefore$$  $$dt\over dx$$ = $$1\over 2at$$]

$$\implies$$ $$d^2y\over dx^2$$  = -$$1\over 2at^3$$

Example : If x = $$acos^3\theta$$ , y = $$asin^3\theta$$ , find $$d^2y\over dx^2$$. Also find its value at $$\theta$$ = $$\pi\over 6$$.

Solution : We have,

x = $$acos^3\theta$$ and y = $$asin^3\theta$$

Differentiating both sides with respect to $$\theta$$,

$$\therefore$$ $$dx\over d\theta$$ = $$-3acos^2\theta sin\theta$$ and $$dy\over d\theta$$ = $$3asin^2\theta cos\theta$$               …….(i)

So, $$dy\over dx$$ = $$dy/d\theta\over dx/d\theta$$ = $$3asin^2\theta cos\theta\over {-3acos^2\theta sin\theta}$$ = $$-tan\theta$$

Differentiating both sides with respect to x, we obtain

$$d^2y\over dx^2$$ = $$d\over dx$$($$-tan\theta$$)

= $$sec^2\theta$$$$d\theta\over dx$$

from (i), [$$dx\over d\theta$$ = $$-3acos^2\theta sin\theta$$ $$\therefore$$  $$d\theta\over dx$$ = $$1\over {-3acos^2\theta sin\theta}$$]

$$d^2y\over dx^2$$ = -$$sec^2\theta$$$$\times$$$$1\over {-3acos^2\theta sin\theta}$$

$$\implies$$ $$d^2y\over dx^2$$ = $$1\over 3a$$ $$sec^4\theta$$$$cosec\theta$$

$$\therefore$$ At $$\theta$$ = $$\pi\over 6$$ , $$d^2y\over dx^2$$  = $$1\over 3a$$ $$sec^4{\pi\over 4}$$$$cosec{\pi\over 6}$$

= $$1\over 3a$$ $$\times$$ $$(2\over \sqrt{3})^4$$ $$\times$$ 2 = $$32\over 27a$$