Question

Finddy/dxandd2y/dx2.x = t2 + 7, y = t2 + 5tdydx = d2ydx2 = For which values of t is the curve concave upward? (Enter your answer using interval notation.)=

Answer 1

`\begin{cases}x(t)=t^2+7\\y(t)=t^2+5t\end{cases}`

By the chain rule,

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

We have

`(\mathrm dx)/(\mathrm dt)=2t`

`(\mathrm dy)/(\mathrm dt)=2t+5`

`\implies(\mathrm dy)/(\mathrm dx)=(2t+5)/(2t)=\boxed{1+\frac5{2t}}`

Notice that this derivative is a function of
`t`. Let
`f(t)=(\mathrm dy)/(\mathrm dx)`. Then by the chain rule,

`(\mathrm d^2y)/(\mathrm dx^2)=(\mathrm df)/(\mathrm dx)=(\mathrm df)/(\mathrm dt)(\mathrm dt)/(\mathrm dx)=((\mathrm df)/(\mathrm dt))/((\mathrm dx)/(\mathrm dt))`

We have

`(\mathrm df)/(\mathrm dt)=-\frac5{2t^2}`

`\implies(\mathrm d^2y)/(\mathrm dx^2)=\frac{-\frac5{2t^2}}{2t}=\boxed{-\frac5{4t^3}}`

The curve is concave upward wherever the second derivative is positive. This happens whenever
`t^3<0`, or
`\boxed{t<0}`.