How much sand is begin removed: use the formula R(t)=2+5sin(4piet/25)
nInt(2+5sin(4piet/25)),t,0,6) = 31.815931 cubic yards of sand [exactly how you input it into a TI-89 calc.]
(b) Write an expression for Y(t), the total number of cubic yards of sand on the beach at time t.
At time t=0, the beach contains 2500 cubic yards of sand.
Create a new formula with the formulas given.
Y(t)=(15t/(1+3t))-2+5sin(4piet/25) dt +2500 (add the 2500)
(c) Find the rate at which the total amount of sand on the beach is changing at time t=4.
find the derv. of y(t)
Y'(t)=(15t/(1+3t))-2+5sin(4piet/25)
plug in 4 for t
Y'(4)=(15(4)/(1+3(4)))-2+5sin(4pie(4)/25)
4.6153846- 6.5241353
= -1.908yard^3 per hour
Y'(t)=(15t/(1+3t))-2+5sin(4piet/25)
plug in 4 for t
Y'(4)=(15(4)/(1+3(4)))-2+5sin(4pie(4)/25)
4.6153846- 6.5241353
= -1.908yard^3 per 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.
need to find where t is a minimum...
equal Y'(t)= 0 to find critical points.
nDeriv or Y(t)
in the interval of 0 to 6...t= 5.117
min value at 2492
im not sure if that's right though...
, at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.need to find where t is a minimum...
equal Y'(t)= 0 to find critical points.
nDeriv or Y(t)
in the interval of 0 to 6...t= 5.117
min value at 2492
im not sure if that's right though...


