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

Question

Charles Anctil · Accepted Answer

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

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 #

William Abdalla · Answer

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

# 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.!

Christopher Agresta · Answer

undefined To find the second derivative for the implicit equation $x^2 + y^2 = a^2$, where $a$ is a constant, follow these steps:

1. Differentiate both sides of the equation with respect to $x$.
2. Solve for $y'$, the first derivative.
3. Differentiate the expression obtained in step 2 with respect to $x$ again.
4. Solve for $y''$, the second derivative.

Here's the breakdown:

1. $2x + 2yy' = 0$
2. $y' = -\frac{x}{y}$
3. Differentiate $y'$ with respect to $x$ again.
4. $y'' = -\frac{y - xy''}{y^2}$

That's the expression for the second derivative, $y''$, in terms of $x$, $y$, and $y''$.