#### (a)

The average rate of change is zero in hte intervals of $0< x < pi, pi < x < 2pi$.

#### (b)

The average rate of change is negative in the intervals of $-dfrac{pi}{2} < x < dfrac{pi}{2}, dfrac{3pi}{2} < x <dfrac{5pi}{2}$.

#### (c)

The average rate of change is positive in the intervals of $dfrac{pi}{2} < x < dfrac{3pi}{2}, dfrac{5pi}{2} < x < 3pi$.

#### (a)

Two points where instantaneous rate of change is zero are $x=dfrac{pi}{4}, x=dfrac{5pi}{4}$.

#### (b)

Two points where the instantaneous rate of change is a negative value are $x=dfrac{pi}{2}, x=dfrac{5pi}{2}$.

#### (c)

Two points where the instantaneous rate of change is positive value are $x=0, x=2pi$

Average rate ofchange for the interval $2 leq x leq 5 :$

$left|dfrac{0-0}{5-2} right|=left|dfrac{0}{3} right|=0$

#### (a)

$0 leq x leq dfrac{pi}{2}$

Average rate of change$=dfrac{2.73-2}{dfrac{pi}{2}-0}=dfrac{0.73}{1.57}=0.465$

#### (b)

$dfrac{pi}{6}leq x leq dfrac{pi}{2}$

Average rate of chnage$=dfrac{2.73-2.73}{dfrac{pi}{2}-dfrac{pi}{6}}=dfrac{0}{1.047}=0$

#### (c)

$dfrac{pi}{3} leq x leq dfrac{pi}{2}$

Average rate of change$=dfrac{2.73-3}{dfrac{pi}{2}-dfrac{pi}{3}}=dfrac{-0.27}{0.5236}=-0.5157$

#### (d)

$dfrac{pi}{2} leq x leq dfrac{5pi}{4}$

Average rate of change$=dfrac{-0.932-2.73}{dfrac{5pi}{4}-dfrac{pi}{2}}=dfrac{-3.662}{2.356}=-1.554$

#### (a)

The average rate of change is zero in the intervals of $0 < x < dfrac{pi}{2}$, $pi < x < dfrac{3pi}{2}$.

#### (b)

The average rate of change is negative in the intervals of $0 < x < dfrac{pi}{4}, pi < x < dfrac{5pi}{4}$

#### (c)

The average rate of change is positive in the intervals of $dfrac{pi}{4} < x < dfrac{pi}{2}, dfrac{5pi}{4} < x <dfrac{3pi}{2}$.

#### (a)

Two points where the instantaneous rate of change is zero are $x=dfrac{1}{4}, x=dfrac{3}{4}$.

#### (b)

Two points where the instantaneous rate of change is a negative value are $x=0, x=1$.

#### (c)

Two points where the instantaneous rate of change is a positive value are $x=dfrac{1}{2}, x=dfrac{3}{2}$.

#### (a)

$y=6cos(3x)+2$ for $dfrac{pi}{4}leq x leq pi$

For $x=dfrac{pi}{4}$,

$y=6cos left( 3left(dfrac{pi}{4} right)right)+2$

$y=-2.2426$

For $x=pi$,

$y=6cos (3(pi))+2$

$y=-4$

Average rate of change$=dfrac{-2.2426-(-4)}{dfrac{pi}{4}-pi}=dfrac{1.7574}{-2.3562}=-0.7459$

#### (b)

$y=-5sinleft(dfrac{1}{2}x right)-9$ for $dfrac{pi}{4}leq x leq pi$

For $x=dfrac{pi}{4}$,

$y=-5sinleft(dfrac{1}{2}left( dfrac{pi}{4}right) right)-9$

$y=-10.9134$

For $x=pi$,

$y=-5sinleft(dfrac{1}{2}pi right)-9$

$y=-14$

Average rate of change$=dfrac{-10.9134-(-14)}{dfrac{pi}{4}-pi}=dfrac{3.0866}{-2.3562}=-1.310$

#### (c)

$y=dfrac{1}{4}cos(8x)+6$ for $dfrac{pi}{4} leq x leq pi$

For $x=dfrac{pi}{4}$,

$y=dfrac{1}{4}cosleft(8left( dfrac{pi}{4}right) right)+6$

$y=6.25$

For $x=pi$,

$y=dfrac{1}{4}cos (8pi)+6$

$6.25$

Average rate of change$=dfrac{6.25-6.25}{dfrac{0}{-2.3562}}=0$

The trip is at its minimum height at $t=0$. Normally the sine function is $0$ at $0$, so the function in this case is translated to the right by $dfrac{1}{4}$ of its period. The propeller makes $200$ revolutions per second, so the period is $dfrac{1}{200}$. The amplitude of the function is the length of the propeller, which is positive. Assume that it is $1$ m for this exercise.
Then the function that describes the height of the tip of the propeller is $h(t)=sin(400pi(t-dfrac{1}{800}))$.
The graph of this function is shown below.

$t=dfrac{1}{300}=0.0033$

From the graph, it is clear that the instantaneous rate of change at $t=dfrac{1}{300}$ is negative.

#### (a)

The axis is at $20.2$.So, the equation of the axis is $y=20.2$ and the amplitude is $4.5$.

$k=dfrac{2pi}{24}=dfrac{pi}{12}$

$R(t)=4.5cosleft( dfrac{pi}{12}tright)+20.2$

#### (b)

fastest: $t=6 months, t=18 months, t=30 months, t=42 months$;

slowest: $t=0 months, t=12 months, t=24 months, t=36 months, t=48 months$

#### (c)

For $5leq tleq7$

$t=5$

$R(5)=4.5cosleft( dfrac{pi}{12}(5)right)+20.2$

$R(5)=21.364$

$t=7$

$R(7)=4.5cosleft(dfrac{pi}{12}(7) right)+20.2$

$R(7)=19.035$

Use the points $(5, 21.364)$ and $(7, 19.035)$.

$dfrac{19.035-21.364}{7-5}=dfrac{-2.329}{2}=-1.164$

$=1.164$ mice per owl/s

#### (a)

The instantaneous rate of change appears to be at its greatest at $12$ hours.

i) For $11 leq t leq 13$

$t=11$

$y=0.5 sinleft(dfrac{pi}{6}(11) right)+4$

$y=3.75$

$t=13$

$y=0.5sinleft(dfrac{pi}{6}(13) right)+4$

$y=4.25$

Use the points $(11, 3.75)$ and $(13, 4.25)$.

$dfrac{4.25-3.75}{13-11}=dfrac{0.5}{2}=0.25 t/h$

ii) For $11.5leq t leq 12.5$

$t=11.5$

$y=0.5sinleft(dfrac{pi}{6}(11.5) right)+4$

$y=3.8706$

$t=12.5$

$y=0.5sinleft(dfrac{pi}{6}(12.5) right)+4$

$y=4.1294$

Use the points $(11.5, 3.8706)$ and $(12.5, 4.1294)$.

$dfrac{4.1294-3.8706}{12.5-11.5}=0.2588$ t/h

iii) For $11.75leq t leq 12.25$

$t=11.75$

$y=0.5sinleft(dfrac{pi}{6} (11.75)right)+4$

$y=3.9347$

$t=12.25$

$y=0.5sinleft(dfrac{pi}{6}(12.25) right)+4$

$y=4.0653$

Use the points $(11.75, 3.9347)$ and $(12.25, 4.0653)$.

$dfrac{4.0653-3.9347}{12.25-11.75}=dfrac{0.1306}{0.5}=0.2612$ t/h

#### (b)

The estimate calculated in part iii) is most accurate. The smaller the interval, the more accurate the estimate.

#### (b)

Half od one cycle

#### (c)

$dfrac{-7.2-7.2}{1-0}=-14.4$ cm/s

#### (d)

The bob is moving the fastest when it passes throug its rest position. You can tell because the images of the balls are farthest apart at this point.

#### (e)

The pendulum’s rest poition is halfway between the maximum and minimum values on the graph.
Therefore, at this point, the pendulum’s instantaneous rate of change is at its maximum.

$h(t)=sinleft(dfrac{pi}{5}t right)$

#### (a)

For $0leq t leq 5$,

$h(0)=sinleft(dfrac{pi}{5}(0) right)=0$

$h(5)=sinleft(dfrac{pi}{5}(5) right)=0$

$dfrac{0-0}{5-0}=0$

#### (b)

For $5.5 leq t leq 6.5$

$t=5.5$

$h(5.5)=sinleft(dfrac{pi}{5}(5.5) right)$

$h(5.5)=-0.309$

$h(6.5)=sinleft( dfrac{pi}{5}(6.5)right)$

$h(6.5)=-0.809$

Use the points $(5.5, -0.309)$ and $(6.5, -0.809)$.

$dfrac{-0.809-(-0.309)}{6.5-5.5}=dfrac{-0.5}{1}=-0.5$ m/s

#### (b)

When $t=0, theta=0.$ When $t=1, theta=0.2$.The average rate of change is $0.2$ radians/s.

#### (c)

Answers may vary. For example, from the graph, it appears that the instantaneous rate of at $t=1.5$ is about $-dfrac{2}{3}$ radians/s.

#### (d)

The pendulum speed seems to be the greatest for $t=0, 2, 4, 6,$and $8$.

Answers may vary. For example, for $x=0$, the instantaneous rate of change of $f(x)=sin x$ is approximately $0.9003$, while the instantaneous rate of change $f(x)=3sin x$ is approximately $2.7009$.
(The interval $-dfrac{pi}{4}< x <dfrac{pi}{4}$was used.)Therefore, the instantaneous rate of change of $f(x)=3sin x$is at its maximum tree times more than the instantaneous rate of change od $f(x)=sin x$.However, there are points where the instantaneous rate of change is the same for the two functions. For example, at $x=dfrac{pi}{2}$, it is $0$ for both functions.

#### (a)

By examining the graph of $f(x)=sin x$, it appears that the instantaneous rate of change at the given values of $x$ are $-1, 0, 1, 0,$ and $-1$.

#### (b)

The function is $f(x)=cos x$. Based on this information, the derivative of $f(x)=sin x$ is $cos x$.

#### (a)

By examining the graph of $f(x)=cos x,$ it appears that the instantaneous rate of change at the given values of $x$ are $0, 1, 0, -1,$ and $0$.

#### (b)

The function is $f(x)=-sin x$. Based on this information, the derivative of $f(x)=cos x$ is $-sin x$.