fbpx
calculus online - Free exercises and solutions to help you succeed!

Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6511

Exercise

Given the differentiable function (with parameter a)

y=f(x-at)+g(x+at)

Prove the equation

y''_{tt}=a^2y''_{xx}

Proof

Define

u=x-at

v=x+at

We get the function

y=f(u)+g(v)

And the internal functions

u(x,t)=x-at

v(y,t)=x+at

We will use the chain rule to calculate the first and second order partial derivatives of y.

y'_t=f'_u\cdot u'_t+g'_v\cdot v'_t=

=f'_u\cdot (-a)+g'_v\cdot a

=-af'_u+ag'_v

 

y''_{tt}=-a\cdot f''_u\cdot u'_t+ag''_v\cdot v'_t=

=-af''_u\cdot (-a)+ag''_v\cdot a=

=a^2f''_u+a^2g''_v

 

y'_x=f'_u\cdot u'_x+g'_v\cdot v'_x=

=f'_u\cdot 1+g'_v\cdot 1=

=f'_u+g'_v

 

y''_{xx}=f''_u\cdot u'_x+g''_v\cdot v'_x

f''_u\cdot 1+g''_v\cdot 1=

=f''_u+g''_v

We will put the second order partial derivatives in the left side of the equation we need to prove.

y''_{tt}=a^2f''_u+a^2g''_v=

=a^2(f''_u+g''_v)=

=a^2y''_{xx}

We got to the right side of the equation as required.

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply