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

blue4882

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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 :D

thank you
 
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 f(t)=t2+14t+7\displaystyle f(t)\:=\:t^2\,+\,14t\,+\,7
Find the instantaneous rate of change of f(t)\displaystyle f(t) with respect to t\displaystyle t at t=2\displaystyle t\,=\,2

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

f(t+h)  =  (t+h)2+14(t+h)+7  =  t2+2th+h2+14t+14h+7\displaystyle f(t + h)\;=\;(t\,+\,h)^2\,+\,14(t\,+\,h)\,+\,7\;=\;t^2\,+\,2th\,+\,h^2\,+\,14t\,+\,14h\,+\,7

f(t+h)f(t)  =  (t2+2th+h2+14t+14h+7)(t2+14t+7)  =  2th+h2+14h\displaystyle f(t\,+\,h)\,-\,f(t)\;=\;(t^2\,+\,2th\,+\,h^2\,+\,14t\,+\,14h\,+\,7)\,-\,(t^2\,+\,14t\,+\,7)\;=\;2th\,+\,h^2\,+\,14h

f(t+h)f(t)h  =  2th+h2+14hh  =  h(2t+h+14)h  =  2t+h+14\displaystyle \frac{f(t\,+\,h)\,-\,f(t)}{h}\;=\;\frac{2th\,+\,h^2\,+\,14h}{h}\;=\;\frac{h(2t\,+\,h\,+\,14)}{h}\;=\;2t\,+\,h\,+\,14

f(t)  =  limh0(2t+h+14)=2t+14\displaystyle f'(t) \;= \;\lim_{h\to0}(2t\,+\,h\,+\,14) \:=\:2t\,+\,14

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

 
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