The tide removes sqnd at a rate modeled by the function R, given by
R(t)= 2+5sin(4pit/25)
A pumping station adds sand to the beach at a rate modeled by the function S, given by
S(t)= 15t/(1+3t)
Both R(t) and S(t) have units of cubic yards per our and t is measured in hours for 0%26lt;-t%26lt;-6. At time t=0, the beach contains cubic yards of sand.
a) How much sand will the tide remove from the beach during this 6-hour period? units?
b) Write an expression for Y(t),the total number of cubic yards of sand on the beach at time t.
c) Find the rate at which the total amount of sand on the beach is changing at time t=4.
d) For 0%26lt;-t%26lt;-6, at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify
Calculus!!!! tide removes sand at a rate of . . HELP!!!!!?
In the small time dt, the sand removed is R(t)*dt, and the sand added is S(t)*dt. Thus, the total vatiation is (sand added) less (sand removed), thus
dm = (S(t)-R(t)) * dt
To find the total sand quantity change, you have to add all the individual sand changes, from the initial moment t=0 to the current time, thus do the integration
Integrate[dm] = Integrate[(S(t)-R(t)) dt, for t=0 to t]
This is the same as adding the total sand added
Integrate[S(t) dt, from t=0 to t]
less the total sand removed
Integrate[R(t) dt, from t=0 to t]
The two are the same, due to properties of the integrals. The first integral is a polynomial, the second is a sine; both integrate straightforward
Calculus!!!! tide removes sand at a rate of . . HELP!!!!!?
a) sum the amount of sand removed:
鈭?+5sin(4蟺t/25) dt = 2t - 25/(4蟺) * cos(4蟺 t/25) |t=0..6
