Question

Find dy/dx and d2y/dx2, and find the slope and concavity (if possible) at the given value of the parameter. (If an answer does not exist, enter DNE.) Parametric Equations Point x = t , y = 7t − 2 t = 9

Answer 1

By the chain rule,

`(\mathrm dy)/(\mathrm dx)=(\mathrm dy)/(\mathrm dt)(\mathrm dt)/(\mathrm dx)`

Then for all
`t` the first derivative has a value of 7.

By the product rule,

`(\mathrm d^2y)/(\mathrm dx^2)=(\mathrm d)/(\mathrm dx)\left[(\mathrm dy)/(\mathrm dt)(\mathrm dt)/(\mathrm dx)\right]=(\mathrm d\left((\mathrm dy)/(\mathrm dt)\right))/(\mathrm dx)(\mathrm dt)/(\mathrm dx)+(\mathrm dy)/(\mathrm dt)(\mathrm d^2t)/(\mathrm dx^2)`

but
`t=x\implies(\mathrm dt)/(\mathrm dx)=1\implies(\mathrm d^2t)/(\mathrm dx^2)=0`, so we're left with

`(\mathrm d^2y)/(\mathrm dx^2)=(\mathrm d\left((\mathrm dy)/(\mathrm dt)\right))/(\mathrm dx)`

By the chain rule,

`(\mathrm d^2y)/(\mathrm dx^2)=(\mathrm d\left((\mathrm dy)/(\mathrm dt)\right))/(\mathrm dx)=(\mathrm d\left((\mathrm dy)/(\mathrm dt)\right))/(\mathrm dt)(\mathrm dt)/(\mathrm dx)=(\mathrm d^2y)/(\mathrm dt^2)`

but
`y=7t-2\implies(\mathrm dy)/(\mathrm dt)=7\implies(\mathrm d^2y)/(\mathrm dt^2)=0` so the second derivative is 0 for all
`t`.