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Calculus 1 : Functions

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Example Questions

Example Question #1 : How To Find Prediction Models

Determine the tangent line to $y=3e^{-x}$ at $x=ln(e^{-2})$ , and use the tangent line to approximate the value at x=ln(e^3) .

Possible Answers:

e^2

-6e^2

-12e^2

12e^2

Correct answer:

-12e^2

Explanation:

First recall that

$ln(e^{-2})=-2$

To find the tangent line of $y=3e^{-x}$ at x=-2 , we first determine the slope of $y=3e^{-x}$ . To do so, we must find its derivative.

Recall that derivatives of exponential functions involving are given as:

$\frac{\mathrm{d} }{\mathrm{d} x}Ae^{g(x)}= A*g'(x)e^{g(x)}$ , where is a constant and g(x) is any function of

In our case, A=3, g(x)=-x, g'(x)=-1 ,.

$\frac{d}{dx}3e^{-x}= -3e^{-x}$

At x=-2 ,

$\frac{\mathrm{d} y}{\mathrm{d} x}=-3e^{-(-2)}}= -3e^2=m$ , where is the slope of the tangent line.

To use point-slope form, we need to know the value of the original function at x=-2 ,

$y=3e^{-x}$

$y(-2)=3e^{-(-2)}=3e^2$

Therefore,

$(y_{appx}-3e^2)=-3e^2(x+2)$

At x=ln(e^3)=3 ,

$(y_{appx}-3e^2)=-3e^2(3+2)$

$y_{appx}=-15e^2+3e^2=-12e^2$

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Example Question #2691 : Functions

Suppose a blood cell increases proportionally to the present amount. If there were blood cells to begin with, and blood cells are present after hours, what is the growth constant?

Possible Answers:

$\frac{3}{2}$

$\frac{1}{2}ln\left(\frac{3}{2}\right)$

$ln\left(\frac{2}{3}\right)$

$ln\left(\frac{3}{2}\right)$

$\frac{2}{3}$

Correct answer:

$\frac{1}{2}ln\left(\frac{3}{2}\right)$

Explanation:

The population size P(t) after some time is given by:

P(t)= P_ie^k^t

where P_i is the initial population.

At the start, there were 30 blood cells.

P_i=P(0)=30

Substitute this value into the given formula.

P(t)= 30e^k^t

After 2 hours, 45 blood cells were present. Write this in mathematical form.

P(2)=45

Substitute this into P(t)= 30e^k^t , and solve for .

45= 30e^k^(^2^)

$\frac{3}{2}= e^2^k$

$ln\left(\frac{3}{2}\right)=ln(e^2^k)$

$2k=ln\left(\frac{3}{2}\right)$

$k=\frac{1}{2}ln\left(\frac{3}{2}\right)$

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Example Question #1 : Constant Of Proportionality

Given any linear function y=mx+b , determine the direct constant of proportionality

Possible Answers:

$\frac{m}{b}$

m/b

$\frac{m-b}{m}$

Correct answer:

Explanation:

Direct constant of proportionality for any given function y, between any x values, is given by

$r=\frac{\Delta y}{\Delta x}$ , where is the direction constant of proportionality

In the case of a linear function

$\frac{\Delta y}{\Delta x}$ is the same thing as the slope.

Therefore, the constant of proportionality is

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Example Question #1 : How To Find Constant Of Proportionality Of Rate

Find the direct constant of proportionality of x^2 from to 2.5 .

Possible Answers:

2.5

3.5

2.5

Correct answer:

3.5

Explanation:

To determine the direct constant of proportionality, we determine the rate of change from x=1 and x=2.5 for x^2 .

Rate of change is determined by

$r=\frac{\Delta y}{\Delta x}$ .

In our case, y=x^2 between x=1 and x=2.5 , the rate of change is

$r=\frac{2.5^2-1}{2.5-1}=\frac{5.25}{1.5}=3.5$ .

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Example Question #3721 : Calculus

Find the direct constant of proportionality of y=ln(3x) from x=2 to x=4 .

Possible Answers:

$r= \frac{6}{2ln(2)}$

$r= \frac{ln(2)}{2}$

r= ln(12)-ln(6)

$r= \frac{2}{ln(2)}$

Correct answer:

$r= \frac{ln(2)}{2}$

Explanation:

Direct constant of proportionality is given by

$r=\frac{\Delta y}{\Delta x}$ .

Since y=ln(3x) and $\Delta x= 4-2=2$

$r= \frac{\Delta y}{\Delta x}=\frac{ln(3*4)-ln(3*2)}{2}= \frac{ln(2)}{2}$

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Example Question #1 : Exponential Growth Applications

Suppose a population of bacteria increases from 100 to 1000 in $30\: min$ . What is the constant of growth?

Possible Answers:

None of these

$\frac{\ln(10)}{30}$

$e^{30}$

$\ln(10)$

Correct answer:

$\frac{\ln(10)}{30}$

Explanation:

The equation for population growth is given by $P=Ae^{kt}$ . is the population, is the intial value, is time, and is the growth constant. We can plug in the values we know at time t=0 and solve for .

$100=Ae^{0k}=A$

Now that we solved for , we can plug in what we know for time t=30 and solve for .

$1000=100e^{30k}$

$e^{30k}=10$

$30k=\ln(10)$

$k=\frac{\ln(10)}{30}$

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Example Question #1 : How To Find Constant Of Proportionality Of Rate

A population of deer grew from 50 to 200 in 7 years. What is the growth constant for this population?

Possible Answers:

$\frac{4}{7}$

$\frac{\ln(4)}{7}$

$\ln(7)$

$\ln(4)$

None of these

Correct answer:

$\frac{\ln(4)}{7}$

Explanation:

The equation for population growth is given by $P=Ae^{kt}$ . P is the population, is the intial value, is time, and is the growth constant. We can plug in the values we know at time t=0 and solve for .

$50=Ae^{0k}=A$

Now that we have solved for we can solve for at t=7

$200=50e^{7k}$

$e^{7k}=4$

$7k=\ln(4)$

$k=\frac{\ln(4)}{7}$

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Example Question #22 : Exponential Functions

A population of mice has 200 mice. After 6 weeks, there are 1600 mice in the population. What is the constant of growth?

Possible Answers:

$\frac{\ln(6)}{8}$

$\ln(8)$

$\frac{\ln(6)}{6}$

$\frac{\ln(8)}{6}$

Correct answer:

$\frac{\ln(8)}{6}$

Explanation:

The equation for population growth is given by $P=Ae^{kt}$ . is the population, is the intial value, is time, and is the growth constant. We can plug in the values we know at time t=0 and solve for .

$200=Ae^{0k}=A$

Now that we have we can solve for at t=6 .

$1600=200e^{6k}$

$e^{6k}=8$

$6k=\ln(8)$

$k=\frac{\ln(8)}{6}$

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Example Question #4 : Constant Of Proportionality

Find the direct constant of proportionality of y=3sec(x) from x=0 to $x=\frac{\pi}{3}$ .

Possible Answers:

$k=\frac{9}{\pi}$

$k=-\frac{\pi}{9}$

$k=\frac{\pi}{9}$

is undefined

Correct answer:

$k=\frac{9}{\pi}$

Explanation:

Direct constant of proportionality is given by

$k=\frac{\Delta y}{\Delta x}$ , where $\Delta y$ is the change in the position and $\Delta x$ is the change in the position.

Since y=3sec(x) , and we're going from x=0 to $x=\frac{\pi}{3}$

$\Delta y=3sec(\frac{\pi}{3})-3sec(0)=6-3=3$

$\Delta x=\frac{\pi}{3}-0=\frac{\pi}{3}$

$k=\frac{3}{\frac{\pi}{3}}=\frac{9}{\pi}$

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Example Question #1 : How To Find Constant Of Proportionality Of Rate

The rate of decrease of the dwindling wolf population of Zion National Park is proportional to the population. The population decreased by 7 percent between 2009 and 2011. What is the constant of proportionality?

Possible Answers:

0.130

-0.036

-0.044

-0.011

-0.009

Correct answer:

-0.036

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population decreased by 7 percent between 2009 and 2011, we can solve for this constant of proportionality:

$(1-0.07)y_0=y_0e^{k(2011-2009)}$

$0.93=e^{2k}$

2k=ln(0.93)

$k=\frac{ln(0.93)}{2}=-0.036$

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : How To Find Prediction Models

Example Question #2691 : Functions

Example Question #1 : Constant Of Proportionality

Example Question #1 : How To Find Constant Of Proportionality Of Rate

Example Question #3721 : Calculus

Example Question #1 : Exponential Growth Applications

Example Question #1 : How To Find Constant Of Proportionality Of Rate

Example Question #22 : Exponential Functions

Example Question #4 : Constant Of Proportionality

Example Question #1 : How To Find Constant Of Proportionality Of Rate

All Calculus 1 Resources

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