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