course Mth 173
The depth of water in a certain uniform cylinder is given by the depth vs. clock time function y = .019 t2 + -1.2 t + 50. What is the average rate at which depth changes between clock times t = 7.4 and t = 14.8?
• What is the clock time halfway between t = 7.4 and t = 14.8, and what is the rate of depth change at this instant?
• What function represents the rate r of depth change at clock time t?
• What is the value of this function at the clock time halfway between t = 7.4 and t = 14.8?
If the rate of depth change is given by dy/dt = .25 t + -1.5 then how much depth change will there be between clock times t = 7.4 and t = 14.8?
• Give the function that represents the depth.
• Give the specific function corresponding to depth 190 at clock time t = 0
1. y= .038(7.4)-1.2+.019(14.8) =.2812-1.2+.2818=-.6376
To get the average rate of change of depth with respect to clock time, you need to evaluate that depth function at the two clock times. Then find ave rate = (change in depth) / (change in clock time).
You appear to be evaluating the rate function (which you incidentally have right), which however isn't appropriate to the first question.
1. Clock time half way between 7.4 and 14.8 = (14.8-7.4) = 7.4(diff of two times)/2(to get half)= 3.7(half time between 14.8&7.4)+7.4(Starting time)=11.1(half way time).
• y’(t)=2(.019)t+(-1.2) =.038t-1.2
• rate = y’(11.1)=.038(11.1)-1.2 = .4218-1.2= -.7782
2. y’= 2at+b
3. y’(11.1)=.038(11.1)-1.2 = .4218-1.2= -.7782
4. For
a. dy/dt = .25t+-1.5
b. y=.125t^2-1.5b+c
c. y(7.4)=.125(7.4)^2-1.5(7.4)+c = 6.845 -11.1+c =-4.255+c
d. y(14.8)=.125(14.8)^2-1.5(7.4)+c = 27.38 -22.2+c =5.18+c
5. y=.125(t)^2-1.5(t)+c
6. 190=.125(0)^2-1.5(0)+c= c, so c=0
If 190=.125(0)^2-1.5(0)+c= c, then c = 190.
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You got most of these questions. See my notes, above and below.
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