Coordinated Calculus

1. On \([0.4,0.8]\text{,}\) the average velocity is \(AV_{[0.4,0.8]} = \frac{s(0.8)-s(0.4)}{0.8-0.4}\) ft/sec.

2. Remember that the slope of a line can be found by taking rise over run. In this context, the slope is found by computing change in \(s\) over change in \(t\text{.}\)

3. Overall, the curve \(y=s(t)\) is a parabola; how does it look up close on a very small interval?

4. Instantaneous velocity can be approximated by average velocity on a very small interval.

1. \(AV_{[0.4,0.8]} = 12.8\) ft/sec; \(AV_{[0.7,0.8]} = 8\) ft/sec; the other average velocities are, respectively, 6.56, 6.416, 0, 4.8, 6.24, 6.384, all in ft/sec.

2. \(m = 12.8\) is the average velocity of the ball between \(t = 0.4\) and \(t = 0.8\text{.}\)

   

3. Like a straight line with slope about 6.4.

4. About 6.4 feet per second.

1. On \([0.4,0.8]\text{,}\) the average velocity is \(AV_{[0.4,0.8]} = \frac{s(0.8)-s(0.4)}{0.8-0.4} = \frac{63.36-58.24}{0.4} = 12.8\) ft/sec. On \([0.7,0.8]\text{,}\) the average velocity is 8 ft/sec. The other average velocities are, respectively (in the order of the intervals listed in the problem statement), 6.56, 6.416, 0, 4.8, 6.24, 6.384, all measured in feet per second.

2. The slope of the line between \(A=(0.4, s(0.4))\) and \(B=(0.8, s(0.8))\) is \(\frac{s(0.8)-s(0.4)}{0.8-0.4} = 12.8\text{.}\) This is precisely the average velocity of the ball between \(t = 0.4\) and \(t = 0.8\text{,}\) and indeed each of the average velocities computed in (a) can be viewed as the slope of the line joining the points \((a,s(a))\) and \((b,s(b))\text{.}\)

   

3. As we zoom in on the graph of \(y=s(t)\) at the point \((0.5, 60)\text{,}\) the graph begins to look like a straight line. Specifically, it appears to look like a straight line with slope about 6.4.

4. Observe that the average velocity of the ball on the intervals \([0.799,0.8]\) and \([0.8,0.801]\) is 6.416 and 6.384 feet/sec respectively. Hence it appears that the ball's velocity at the instant \(t = 0.8\) should be about 6.4 feet per second.

1. Remember to use the formula for average velocity from above: \(AV_{[a,b]} = \frac{s(b)-s(a)}{b-a}\text{.}\) Think carefully about whether certain quantities are positive or negative.

2. To estimate the instantaneous velocity at \(t = 1.5\text{,}\) consider average velocities on the intervals \([1.499,1.5]\) and \([1.5,1.501]\text{.}\)

3. Think about whether the ball is rising or falling.

4. What is the average velocity of the ball on small intervals that contain \(t = 1\text{?}\)

1. \(AV_{[1.5,2]} = -24\) ft/sec, which is negative; \(AV_{[0,0.5]} = 24\) ft/sec, which is positive.

2. The instantaneous velocity at \(t = 1.5\) is approximately \(-16\) ft/sec; at \(t = 2\text{,}\) the instantaneous velocity is about \(-32\) ft/sec, and \(-16>-32\text{.}\)

3. When the ball is rising, its instantaneous velocity is positive; when the ball is falling, its instantaneous velocity is negative.

4. Zero.

1. \(AV_{[1.5,2]} = \frac{s(2)-s(1.5)}{2-1.5} = -24\) ft/sec. We note that this average velocity is negative, and in fact is the opposite of the average velocity of 24 ft/sec on the interval \([0,0.5]\text{.}\)

2. Since \(AV_{[1.499,1.5]} = -15.984\) and \(AV_{[1.5, 1.501]} = -16.016\text{,}\) it appears that the instantaneous velocity of the ball at \(t = 1.5\) is approximately \(-16\) ft/sec. Similar computations show that at \(t = 2\text{,}\) the instantaneous velocity is about \(-32\) ft/sec. Note that \(-16>-32\text{,}\) so the instantaneous velocity at \(t = 1.5\) is greater because it is less negative. Asking which number is greater is different from asking which number is more negative.

3. When the ball is rising, its instantaneous velocity is positive; when the ball is falling, its instantaneous velocity is negative.

4. Note that \((1,s(1))\) is the vertex of the parabola given by \(s(t)\text{.}\) At this point, the ball is neither rising nor falling. On intervals of the form \([a,1]\text{,}\) where \(a \lt 1\text{,}\) the average velocity of the ball is positive; on intervals of form \([1,b]\text{,}\) where \(b > 1\text{,}\) the average velocity is positive. Hence we expect the instantaneous velocity of the ball at the moment \(t = 1\) to be zero.

1. What formula do we have for the average velocity of a function \(f\) on an interval \([a,b]\text{?}\) Can this be simplified at all with this function and interval? Why is this range of \(h\) values specified?
2. What values of \(h\) do we need to use to get these intervals? Be careful with whether you choose \(h\) to be positive or negative.
3. How might we use the expression we found for average velocity to estimate instantaneous velocity?

1. \(AV_{[0.5, 0.5+h]} = -16 - 16h\text{.}\)
2. Use \(h=0.25\) to find \(AV_{[0.5,0.75]}=-20\) ft/sec, and \(h=-0.1\) to find \(AV_{[0.4,0.5]}=-14.4\) ft/sec.
3. The instantaneous velocity at \(t=0.5\) should be \(-16\) ft/sec.

1. We make the assumptions that \(-0.5 \lt h \lt 0.5\) and \(h \ne 0\) because \(h\) cannot be zero (otherwise there is no interval on which to compute average velocity) and because the function only makes sense on the time interval \(0 \le t \le 1\text{,}\) as this is the duration of time during which the ball is falling. We want to compute and simplify
   \begin{equation*} AV_{[0.5, 0.5+h]} = \frac{s(0.5+h) - s(0.5)}{(0.5+h) - 0.5}\text{.} \end{equation*}
   We start by finding \(s(0.5+h)\text{.}\) To do so, we follow the rule that defines the function \(s\text{.}\)
   \begin{align*} s(0.5+h) \amp = 16 - 16(0.5 + h)^2\\ \amp = 16 - 16(0.25 + h + h^2)\\ \amp = 16 - 4 - 16h - 16h^2\\ \amp = 12 - 16h - 16h^2\text{.} \end{align*}
   Now, returning to our computation of the average velocity, we find that
   \begin{align*} AV_{[0.5, 0.5+h]} \amp = \frac{s(0.5+h) - s(0.5)}{(0.5+h) - 0.5}\\ \amp = \frac{(12 - 16h - 16h^2) - (16 - 16(0.5)^2)}{0.5 + h - 0.5}\\ \amp = \frac{12 - 16h - 16h^2 - 12}{h}\\ \amp = \frac{-16h - 16h^2}{h}\text{.} \end{align*}
   At this point, we note two things: first, the expression for average velocity clearly depends on \(h\text{,}\) which it must, since as \(h\) changes the average velocity will change. Further, we note that since \(h\) can never equal zero, we may remove the common factor of \(h\) from the numerator and denominator. It follows that
   \begin{equation*} AV_{[0.5, 0.5+h]} = -16 - 16h\text{.} \end{equation*}
2. From this expression we can compute the average velocity on \([0.5,0.5+h]\) for any small positive or negative value of \(h\text{.}\) For instance, to obtain the average velocity on \([0.5,0.75]\text{,}\) we let \(h = 0.25\text{,}\) and the average velocity is \(-16 - 16(0.25) = -20\) ft/sec. To get the average velocity on \([0.4, 0.5]\text{,}\) we let \(h = -0.1\text{,}\) and compute the average velocity as
   \begin{equation*} -16 - 16(-0.1) = -14.4\ \text{ft/sec}\text{.} \end{equation*}
3. Using our formula from (a), we can explore what happens to \(AV_{[0.5, 0.5+h]}\) as \(h\) gets closer and closer to zero. As \(h\) approaches zero, \(-16h\) will also approach zero, so it appears that the instantaneous velocity of the ball at \(t = 0.5\) should be \(-16\) ft/sec.

Note that

\(AV_{[2, 2+h]} = -32 - 16h\text{,}\) so \(AV_{[1.5,2]}=-24\) ft/sec. The instantaneous velocity at \(t=2\) is \(-32\) ft/sec.

Observe first that

Next, recall that \(AV_{[2, 2+h]} = \frac{s(2+h) - s(2)}{h}\text{,}\) so

Now, since we assume \(h \ne 0\text{,}\) we can simplify further to find that \(AV_{[2, 2+h]} = -32 - 16h\text{.}\) Setting \(h = -0.5\text{,}\) it follows that \(AV_{[1.5,2]} = -32 + 16(0.5) = -24\) ft/sec. Likewise, as we let \(h\) approach zero, we see that \(-32 - 16h\) will approach \(-32\text{,}\) so the instantaneous velocity at \(t = 2\) appears to be \(-32\) ft/sec. Both results match our earlier work in Example1.42.