A pumping station adds sand to the beach at a rate modeled by the function S, given by
Both R(t) and S(t) have units of cubic yards per hour and t is measured in hours for
. At time t=0, the beach contains 2500 cubic yards of sand.(a) How much sand will the tide remove from the beach during this 6-hour period? Indicate units of measure.
*integral of R(t) from 0 to 6 of R(t) dx
-Since R(t) is the rate at which sand is REMOVED, the following is input into the calculator to figure out the amount of sand that is removed after 6 hours: fnInt(2+5sin(4πt/25),t,0,6) ≈ 31.816 cubic yards of sand
(b) Write an expression for Y(t), the total number of cubic yards of sand on the beach at time t.
-R(t) is used to determine the amount of sand that is removed at t and S(t) is used to determine the amount of sand that is added at t. As a result, you subtract S(t) by R(t) to find the TOTAL number of cubic yards of sand on the beach at time t and 2,500 is added since that is the amount of sand that the beach originally contains: y(t) = fnInt(15t/1+3t)-fnInt(2+5sin(4πt/25))+2,500
(c) Find the rate at which the total amount of sand on the beach is changing at time t=4.
-Since y(t) represents the total number of cubic yards of sand on the beach at time t, y'(t) represents the RATE at which the total number of sand on the beach is changing.
y'(t)=15/(1+3t)^2 - 5cos(4πt/25)(0.05024)
y'(4)≈2.1243 cubic yards/hour
(d) For
, at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.-At t=5.1178, the amount of sand on the beach is at its minimum since both of the rates are equal which means that the same amount of sand is removed as the same amount of sand that is added.
-At t=5.1178, there is a minimum of 4.6943(+2,500) which adds up to 2504.6943 cubic yards of sand
for a and b i got the same answers. but for c and d.. i got something different
ReplyDeletei dont quite understand why ... y'(t)=15/(1+3t)^2 - 5cos(4πt/25)(0.05024) .. the squared part... or is it just a typo?
well i got this
y'(t) = nDeriv((15t/(1+3t)) - (2+5sin(4π t/25) + 2500)
y'(4)= 1.159 cubic yard/hour/hour
so some how.. we did a similar method but we got different answers.
and for part d.. i did it on the yellow calculator.. i set S(t) as y1, R(t) as y2, and y1-y2 as y3. so i just graphed y3. instead of trying to find the critical point, i just calculated the minimum. and i found it to be at x=3.007 .. dont forget to look at your end points 0,6.. but yea. i wonder how i got that answer...
so far.. i've seen your's and stephanie herrera's.. hahah and we all got different answers. either one of us is right.. or we're all wrong!!
:D
Hey for c, I think you made a mistake, but I dont know where, since i got -1.909. It is S(t)-R(t), but maybe its in your math? (:
ReplyDeletefor ppart c would u leave it how it is and not take the derivative
ReplyDeletefor part c, you would use you original equation, or S(t)-R(t), because that will output the rate at which the total number of sand on the beach is changing.
ReplyDelete