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On a typical day, the snow on Pike Mountain melts at a rate modeled by the function

M(t), given by M(t)=π/6sin(πt/12)

A snow maker adds snow at a rate modeled by the function S, given by

S(t)=0.006t^2−0.12t+0.87

Both M and S have units in inches per hour and t is measured in hours for 0≤t≤6. At time t = 0, the mountain has 40 inches of snow.

a. How much snow will melt during the 6 hour period? Indicate units of measure.
b. Write an expression for I(t), the total number of inches of snow at any time t.
c. Find the rate of change of the total amount of snow at time t = 3.

Sagot :

ajdonis

Answer:

a. 2in

b. [tex]l(t)=0.002t^{3}-0.06t^{2}+0.87t+2cos(\frac{\pi}{12}t)+38[/tex]

c. l'(3)=0.1938 in/hr

Step-by-step explanation:

a.

In part a, the problem is asking us to find how much snow will melt during the 6 hour period. In order to do so we need to take the M(t) function and integrate it in the given period of time, so we get:

[tex]\int\limits^6_0 {\frac{\pi}{6}sin(\frac{\pi}{12}t)} \, dt[/tex]

So now we solve:

[tex]\frac{\pi}{6}\int\limits^6_0 {sin(\frac{\pi}{12}t)} \, dt[/tex]

this is a known integral, so we get:

[tex]\frac{\pi}{6}[-\frac{12}{\pi}cos(\frac{\pi}{12}t)]^{6}_{0}[/tex]

we can simplify this so we get:

[tex]-2[cos(\frac{\pi}{12}t)]^{6}_{0}[\tex]

[tex]-2[cos(\frac{\pi}{12}(6))-cos(\frac{\pi}{12}(0))][/tex]

[tex]-2[cos(\frac{\pi}{2})-cos(0)][/tex]

[tex]-2[0-1][/tex]

2 in

b)

For part b they want us to write an expression for l(t), the total number of inches of snow at any time t.

in order to do so we need to find an expression for the rate of change of snow. A snowmaker is adding some snow while another amountn of snow is melting, so this rate of change is found by subtracting the two functions, so we get:

l'(t)=S(t)-M(t)

so:

[tex]l(t)=\int {(0.006t^{2}-0.12t+0.87-\frac{\pi}{6}sin(\frac{\pi}{12}t))} \, dt[/tex]

so we integrate this to get:

[tex]l(t)=\frac{0.006}{3}t^{3}-\frac{0.12}{2}t^{2}+0.87t+2cos(\frac{\pi}{12}t)+C[/tex]

and we can simplify this:

[tex]l(t)=0.002t^{3}-0.006t^{2}+0.87t+2cos(\frac{\pi}{12}t)+C[/tex]

so now we need to find what C is equal to, we can use the fact that l(0)=40in so we get:

[tex]40=0.002(0)^{3}-0.006(0)^{2}+0.87(0)+2cos(\frac{\pi}{12}(0))+C[/tex]

which yields:

40=2+C

so

C=40-2

C=38

So the final function is:

[tex]l(t)=0.002t^{3}-0.06t^{2}+0.87t+2cos(\frac{\pi}{12}t)+38[/tex]

c)

for part c we need to find the rate of change of the total amount of snow at time t=3, so in this case we can use the equation we found previously:

l'(t)=S(t)-M(t)

[tex]l'(t)=(0.006t^{2}-0.12t+0.87-\frac{\pi}{6}sin(\frac{\pi}{12}t))[/tex]

and substitute t=3

[tex]l'(3)=(0.006(3)^{2}-0.12(3)+0.87-\frac{\pi}{6}sin(\frac{\pi}{12}(3)))[/tex]

which yields:

l'(t)=0.1938 in/hr

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