Bamboo belongs to the grass family and is one of the fastest growing plants in the world. (a). A bamboo shoot was $500 \mathrm{~cm}$ tall at $10.00$ a.m. and $515 \mathrm{~cm}$ at $3:00$ p.m. Compute the average growth rate of the bamboo shoot in $\mathrm{cm} / \mathrm{hr}$ over the period of time from $10:00$ a.m. to $3:00$ p.m. (b). Based on the Mean Value Theorem, what can you conclude about the instantaneous growth rate of bamboo measured in millimeters per second between $10:00$ a.m. and $3:00$ p.m.?

Is the statement in the following problem true or false? Give an explanation for your answer.

If $P=4 e^{-0.90 t}$, we say the continuous growth rate of the function is $10 \%$.

An evergreen nursery usually sells a certain shrub after 6 years of growth and shaping. The growth rate during those 6 years is approximated by dh/dt=1.5t+5 where t is the time in years and h is the height in centimeters. The seedlings are 12 centimeters tall when planted (t=0). Find the height after t years.

An evergreen nursery usually sells a certain shrub after 6 years of growth and shaping. The growth rate during those 6 years is approximated by $d h / d t=1.5 t+5$, where $t$ is the time in years and $h$ is the height in centimeters. The seedlings are 12 centimeters tall when planted $(t=0)$.

How tall are the shrubs when they are sold?

From 2 to 13 years, the growth rate for children is generally linear. Choose the letter of the correct answer.

Write the equation from the previous Exercise in slope-intercept form.

(A) $y=2.4 x-100.6$

(B) $y=44 x-217.6$

($C$) $y=5 x+32$

(D) $y=2.4 x+32$

To show that the exponential growth rate can be determined using any two points, let $y_1=C e^{k t_1}$, and $y_2=C e^{k t_2}$. Solve this system of equations for $k$ to show that $k$ can be calculated directly, using $\left(t_1, y_1\right)$ and $\left(t_2, y_2\right)$.

In 2004, the world population was estimated to be 6.4 billion, with an annual growth rate of 1.14%. a. Define input and output variables for this situation. b. Without finding an equation, sketch a graph of this situation for the years 1995 to 2015. c. What one point on the graph do you know for sure? d. Write a function that models this situation. Graph your function on your calculator and describe an appropriate window. e. Use your graph to estimate the population to the nearest tenth of a billion in 1995 and 2015. (Assume a constant growth rate during this period.)

Show that the growth rate k of the exponential function $f ( t ) = P _ { 0 } e ^ { k t },$ with $P_0 > 0,$ can be computed from two values of f by a difference quotient of the form: $k = \frac { \ln f \left( t _ { 2 } \right) - \ln f \left( t _ { 1 } \right) } { t _ { 2 } - t _ { 1 } }.$

From 2 to 13 years, the growth rate for children is generally linear. Choose the letter of the correct answer.

Use the equation in the previous Exercise to find the average height of a 13-year old girl.

(F) $56.3$ in. (G) $63.2$ in. (H) $69.4$ in. (J) $97$ in.

Let $f(t) = Q_0 a^t = Q_0 (1 + r)^t.$ (a) Find the base, a. (b) Find the percentage growth rate, r. f(5) = 75.94 and f(7) = 170.86

From 2 to 13 years, the growth rate for children is generally linear. Choose the letter of the correct answer.

The average height of a 5 -year old girl is 44 inches, and the average growth rate per year is $2.4$ inches. Write an equation in point-slope form that describes the height of a girl $y$ based on her age $x$.

(F) $y-2.4=44(x-5)$

(G) $y-44=2.4(x-5)$

(H) $y-44=5(x-2.4)$

(J) $y-5=2.4(x-44)$

In $1960$ a gallon of milk was around $\$ 0.49$. In $2009$ a gallon of milk was around $\$ 4.59$. Use the formula for compound interest to find the annual growth rate (to the nearest tenth of a percent) for the price of a gallon to milk.

If the growth rate of the number of bacteria at any time t is proportional to the number present at t and doubles in 1 week, how many bacteria can be expected after 2 weeks? After 4 weeks?

At the beginning of this section, we modified the exponentialgrowth equation to include oscillations in the per capita growth rate. Solve the differential equation we obtained, namely, $\frac{d N}{d t}=2(1+\sin (2 \pi t)) N(t)$ with N(0)=5.