The growth rate for the dwarf juniper plant t years after planting is approximated by the function:?

The growth rate for the dwarf juniper plant t years after planting is approximated by the function:








dh = 17.6 t inches per year


dt √17.6t²+1








If a juniper seeding is 4 inches tall when first planted, find a function h(t)=height of the plant after t years.











What is the height of the plant after 5 years???

The growth rate for the dwarf juniper plant t years after planting is approximated by the function:?
I'm assuming


(dh/dt) = (17.6t) / (17.6 t^2 + 1)


Integrating this gives


h = sqrt (17.6 t^2 + 1) + c


Plugging in h(0) = 4 gives c = 3


h(5) = sqrt (17.6(25) + 1) + 3


Answer: h(5) = 24 inches
Reply:after 5 years :





4 inch + integral 17.6 t / √17.6t²+1 dt : between t=0 and t=5





.





integral 17.6 t / √17.6t²+1 dt = sqrt(17.6t^2 + 1)





.now its only a matter of filling in the numbers.
Reply:That's an easy one to integrate


If you don't know how, use the substitution


u = 17.6t^2 + 1





h - h0 = sqrt(17.6t^2 + 1) - 1





I played around with the initial conditions there.


You may be more comfortable writing it in the form


h = sqrt(17.6t^2 + 1) + c





Anyway, when t = 5


h - 4 = sqrt(441) - 1 = 20





h = 24 inches = 2 feet
Reply:dh/dt = 17.6t/(sqrt(17.6t^2 + 1))





= 17.6t(17.6t^2 + 1)^(-1/2)





Integrate dh/dt with respect to t by letting U = (17.6t^2 + 1)


and du = 2*17.6t





I get h(t) = (17.6t^2 + 1)^(1/2) + C





Put in h(0) = (17.6*0^2 + 1)^(1/2) + C and get C = 3.





So, h(5) = (17.6*5^2 + 1)^(1/2) + 3 = (17.6*25 + 1)^(1/2) + 3





or Sqrt(450) + 3 = 3*Sqrt(10) + 3.

arenas