Question

How do you find the second derivative for the implicit equation `x^2+y^2 = a^2`?

Answer 1

`(d^2y)/(dx^2) = -a^2/y^3`

Explanation:

We have:

`x^2+y^2 = a^2`

This represents a circle of radius `a` centred on the origin.

If we differentiate implicitly wrt `x` we get:

`2x + 2ydy/dx = 0`

`:. x + ydy/dx = 0 => dy/dx = -x/y`

Now, differentiating implicitly again, and applying the product rule, we get:

`:. 1 + (y)((d^2y)/(dx^2)) + (dy/dx)(dy/dx) = 0`

`:. 1 + y(d^2y)/(dx^2) + (dy/dx)^2 = 0`

`:. 1 + y(d^2y)/(dx^2) + (-x/y)^2 = 0`

`:. 1 + y(d^2y)/(dx^2) + x^2/y^2 = 0`

`:. y(d^2y)/(dx^2) + (x^2+y^2)/y^2 = 0`

`:. y(d^2y)/(dx^2) + a^2/y^2 = 0`

`:. (d^2y)/(dx^2) = -a^2/y^3`

Answer 2

`(d^2y)/dx^2=-a^2/y^3.`

Explanation:

`x^2+y^2=a^2.`

`:. d/dx(x^2+y^2)=d/dx(a^2)=0.`

`:. d/dx(x^2)+d/dx(y^2)=0.`

`:. 2x+d/dx(y^2)=0.............(ast).`

Here, by the Chain Rule, we see that,

`d/dx(y^2)=d/dy(y^2)*dy/dx=2ydy/dx.`

`:. (ast) rArr 2x+2ydy/dx=0, or, x+ydy/dx=0....(ast_1).`

To get the `2^(nd)` order deri., we rediff. this eqn. w.r.t. `x`, & get,

`1+d/dx(ydy/dx)=0............(star).`

Here, for `d/dx(ydy/dx),` we use the Product Rule :

`d/dx(ydy/dx)=yd/dx(dy/dx)+(dy/dx)(d/dx(y)),`

`=y(d^2y)/dx^2+(dy/dx)(dy/dx), i.e., y(d^2y)/dx^2+(dy/dx)^2.`

`:. (star) rArr 1+y(d^2y)/dx^2+(dy/dx)^2=0......(star_1).`

But, `(ast_1) rArr dy/dx=-x/y.`

`:. (star_1) rArr 1+y(d^2y)/dx^2+(-x/y)^2=0.`

`rArr y^2+y^3(d^2y)/dx^2+x^2=0.`

`rArr y^3(d^2y)/dx^2=-(x^2+y^2), i.e.,`

`because, x^2+y^2=a^2, :., (d^2y)/dx^2=-a^2/y^3.`

Alternatively, the same result can be obtained by diff.ing, w.r.t.

`x`, the eqn. `dy/dx=-x/y.`

Enjoy Maths.!