suppose h(t)=t^2+14t+7. Find the instantaneous rate of

[blue4882]

suppose h(t)=t^2+14t+7. Find the instantaneous rate of change of h(t) with respect to t at t=2

any help would be greatly appreciated

thank you

 

[soroban]

Hello, blue4882!

I assume you are using the Definition of a Derivative:

\(\displaystyle \L\;\;f'(x) \;= \;\lim_{h\to0}\frac{f(x + h)\,-\,f(x)}{h}\)

Suppose \(\displaystyle f(t)\:=\:t^2\,+\,14t\,+\,7\)
Find the instantaneous rate of change of \(\displaystyle f(t)\) with respect to \(\displaystyle t\) at \(\displaystyle t\,=\,2\)

Let's do it one step at a time . . .

\(\displaystyle f(t + h)\;=\;(t\,+\,h)^2\,+\,14(t\,+\,h)\,+\,7\;=\;t^2\,+\,2th\,+\,h^2\,+\,14t\,+\,14h\,+\,7\)

\(\displaystyle f(t\,+\,h)\,-\,f(t)\;=\;(t^2\,+\,2th\,+\,h^2\,+\,14t\,+\,14h\,+\,7)\,-\,(t^2\,+\,14t\,+\,7)\;=\;2th\,+\,h^2\,+\,14h\)

\(\displaystyle \frac{f(t\,+\,h)\,-\,f(t)}{h}\;=\;\frac{2th\,+\,h^2\,+\,14h}{h}\;=\;\frac{h(2t\,+\,h\,+\,14)}{h}\;=\;2t\,+\,h\,+\,14\)

\(\displaystyle f'(t) \;= \;\lim_{h\to0}(2t\,+\,h\,+\,14) \:=\:2t\,+\,14\)

At \(\displaystyle t\,=\,2:\;\;f'(2)\;=\;2(2)\,+\,14\;=\;18\)