Tuesday+February+16+Homework+Calculus

a copy of the worksheet is above. I am anticipating that we will go over problems #1 and #2 in class;

-to find the amount of people at a given time (t=2) we must find the integral over the given time. -Take the integral of R(t) --> [460t^3-168.75t^4]dt (the constant that comes about due to integration can be ignored) -Integral of R(t) over t=0 to t=2 --> [460(2^3)-168.75(2^4)]-([460(0^3)-168.75(0^4)] = 980 people who have entered the auditorium by t=2 hours
 * 3: Josh Rubin
 * 1) 3-A) units for R(t)=people per hour, t=hours

B) To find the maximum slope of R(t) take the derivative, set it equal to zero and find all values at which x=0 then plug those values into the original equation. The value of x that yields the highest number is the maximum rate at which people are entering the auditorium. R(t)'=2760x-2025x^2 2760x-2025x^2=0 roots at x=184/135=1.36296 and at x=0

plugging these roots into the first original equation of R(t)=1380t^2-675t^3 we get R(1.36296)=854.5273 people per hour entering the auditorium R(0)=0

therefore at t= 1.36296 hours, the rate of people entering the auditorium is the highest during the 2 hour period. It cost different prices to make the first meter of cable, the second meter of cable, the third, etc. so the cost of a 50-meter cable would be found by summing the cost of making each of these individual meters! Summation means finding the integral! Therefore the cost of making a cable that is x meters long is as follows: This is actually the answer I got for c. So this is your profit function. You can actually integrate this pretty easily and get a function without an integral in it...and then you need to consider what's going on at your endpoints (what are your endpoints), and what critical values you have.
 * 4: Angus This one is a bit of a struggle. Here is a start:


 * 5: Brian

You can do the same thing to find f(4). Then for the absolute max, you have already looked t the endpoints. Where is f'(x) = 0? These will be your critical values! b. Find f(-4) and f(4) somehow take the anti-derivative of f prime (x) between -4 and 0 the anti-derivative is ((x + 2)^3) / 3 therefore f(-4) would be 72 between 0 and 4 the anti-derivative would be [5e^(-x/3 + 1)] / (-x/3 + 1) - 3x and f(4) would be -22.748

c. take the integral over -4 to 4 and then set the derivative equal to zero