Question

I really need help with this calculus 1 question?

Answer 1

Answer:

a) The rate of change brightness after 't' years

  $\frac{dB}{dt} = 5.0+0.55cos( \frac{2\pi(t) }{4.7})(\frac{2\pi }{4.7})$

b) The rate of increase after five years

$(\frac{dB}{dt})_{t=5} = 0.67688$

Step-by-step explanation:

a) Given B(t) = 5.0+0.55sin($\frac{2\pi(t) }{4.7}$   .....(1)

The rate of change of the brightness after 't' days that is

$\frac{dB}{dt}$

now differentiating equation (1) with respective to 't'

$\frac{dB}{dt} = 5.0+0.55cos( \frac{2\pi(t) }{4.7})\frac{d}{dt} (\frac{2\pi(t) }{4.7})$       { using d/dx(sin x) =cos x}

$\frac{dB}{dt} = 5.0+0.55cos( \frac{2\pi(t) }{4.7})(\frac{2\pi }{4.7})$

b) The rate of increase after five days    

substitute t = 5 in equation dB/d t

$\frac{dB}{dt} = 5.0+0.55cos( \frac{2\pi(5) }{4.7})(\frac{2\pi }{4.7})$

after calculation $(\frac{dB}{dt})_{t=5} = 0.67688$