Let  where f is the function graphed above.
     (a)    Evaluate g(4) and g'(4).
     (b)    Where on the interval [0,5] is g increasing? concave up?
     (c)    How many roots does g have on [0,5]?
     (d)    Rank these numbers from lowest to highest: -1, 0, 1, g(0), g(2), g(4).
     (e)    Is the average value of g'(x) over the interval [0,3] greater than 1? Justify.

Difficulty Rating: 
         Source:   Calculus from Graphical, Numerical, and Symbolic Points of View, Ostebee and Zorn, 1997, p. 376.

---

Solution: (a)    This is an accumulation function, so g gets its area by acumulating area beneath f. Because these are circles (two quarter-circles in this part of the problem), this is easy. Note, however, that the functions begins accumulating at 1, according to the boundaries:

(b)    Remember that g'(x) = f(x), so when f is positive, g is increasing. Thus, g is increasing on (2,5). Similarly, when f '(x) is positive, g is concave up (since f '(x)=g''(x)). Also, f '(x) is positive when f(x) is increasing, so g(x) is concave up on (1,4).

(c)    In order to do this, we should sketch a graph of g(x). I used the below table (which I got by plugging in each x into g(x) and finding the accumlated area like we did in part (a)) to graph g, and graphing it makes it clear that g(x) has two roots, or x-intercepts.

(d)    Once we've done the chart for part (c), this is easy. The order goes: -1, g(0), 0, g(2), 1, g(4).

(e)    A major part of this problem will be approximating the area from x = 0 to 3 on the graph. Up to x = 2 is easy, but that area between x = 2 and x = 3 needs to be approximated. I approximate is as about 1.3 square units. Then, I use the average value formula to get:

It is clear that the average value must be less than one, not greater. In order for the average value to be 3, the area between x = 2 and x = 3 would have to larger than 3! The area between x = 0 and x = 2 is already quite negative, so the area on (2,3) would have to make up for that to ensure that the total is greater than 3, so that once we multiply by (1/3), the average value would be greater than one. Clearly, that small sliver of area between x = 2 and x = 3 cannot be larger than 3 square units, so the answer is no.

Calculus II (BC): Approximating Volume with Tables

I am recycling hurricane lamp glass into very hip fishtanks. One (on its side) is shown above. I
want to approximate the volume of the tank. Yes, I know I could simply fill it with water and then measure the water, but then there'd be no Problem of the Week this week, and you ... you would be devestated. I know it.

I have measured the diameter of the tank at regular intervals along its length, and these measurements are shown below (all in inches):

     (a)    Use these data points to approximate the volume of the tank.

     (b)    Use your calculator to plot these data points and to design a fourth-degree regression polynomial to approximate a function through these points. (Look in your calculator manual if you are not sure what I am talking about.)

     (c)    Use your graphing calculator to approximate the volume using the regression function.

Difficulty Rating: 

Solution: (a)    I used the Trapezoidal Rule to approximate the area beneath the curve: Although there are other methods, I decided to find the average value of this function, knowing that I could find rotational volume very easily with a simple cylinder. Use average value formula (and the fact that we approximate the area to be 45) to get that the average value of the function will be (1/9)(45)=5. So, we are finding the volume of a right cylinder with radius 5 and height 9. Thus, our approximateion for the volume of the fish tank is 706.858 in³.

(b)    On the TI-83, you enter the points in L1 and L2 and use Stat--Calc--QuartReg. Below is the regression curve graphed and its equation.

(c)    Once you get the regression equation and type it into Y1 (again, instructions for the TI-83), you can use fnInt (Math--9) in order to find the definite integral of the area. It turns out to be very close to 45! I then divided by 9 (to get average value again) to calculate volume for a final answer of 706.798 in³.

---