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Calculus 1 : Calculus

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Example Questions

Example Question #141 : Constant Of Proportionality

The rate of change of the number of man-eating spore plants is proportional to the population. The population increased from 3 to 6561 between May and November. What is the constant of proportionality in months^-1?

Possible Answers:

4.23

0.85

3.92

1.28

5.90

Correct answer:

1.28

Explanation:

We're told that the rate of change 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 increased from 3 to 6561 between May and November, we can solve for this constant of proportionality. Represent the months by their number in the calendar:

$6561=3e^{k(11-5)}$

$2187=e^{6k}$

6k=ln(2187)

$k=\frac{ln(2187)}{6}=1.28$

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Example Question #142 : Constant Of Proportionality

The rate of change of the number of yeast cells in a pond is proportional to the population. The population increased from 1,800 to 76,000 between 3pm and 8pm. What is the constant of proportionality in hours^-1?

Possible Answers:

2.13

1.98

0.75

1.02

2.46

Correct answer:

0.75

Explanation:

We're told that the rate of change 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 increased from 1,800 to 76,000 between 3pm and 8pm, we can solve for this constant of proportionality:

$76000=1800e^{k(8-3)}$

$\frac{380}{9}=e^{5k}$

$5k=ln(\frac{380}{9})$

$k=\frac{ln(\frac{380}{9})}{5}=0.75$

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Example Question #143 : Constant Of Proportionality

The rate of change of the number of golems is proportional to the population. The population decreased from 324 to 18 between 897 and 895 BC. What is the constant of proportionality in years^-1?

Possible Answers:

2.75

1.45

-1.45

-3.58

3.58

Correct answer:

-1.45

Explanation:

We're told that the rate of change 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 from 324 to 18 between 897 and 895 BC, we can solve for this constant of proportionality. Remember that BC counts down:

$18=324e^{k(897-895)}$

$\frac{1}{18}=e^{2k}$

$2k=ln(\frac{1}{18})$

$k=\frac{ln(\frac{1}{18})}{2}=-1.45$

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Example Question #144 : Constant Of Proportionality

The rate of change of the number of will-o-wisps is proportional to the population. The population decreased from 9500 to 38 between 3 AM and 6 AM. What is the constant of proportionality in hours^-1?

Possible Answers:

-6.62

-4.42

-1.84

-0.37

-16.56

Correct answer:

-1.84

Explanation:

We're told that the rate of change 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 from 9500 to 38 between 3 AM and 6 AM, we can solve for this constant of proportionality:

$38=9500e^{k(6-3)}$

$\frac{1}{250}=e^{3k}$

$3k=ln(\frac{1}{250})$

$k=\frac{ln(\frac{1}{250})}{3}=-1.84$

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Example Question #3861 : Calculus

The rate of change of the number of active ghosts is proportional to the population. The population decreased from 20,000 to 10 between 4 AM and 5 AM. What is the constant of proportionality in hours^-1?

Possible Answers:

-3.00

-9.90

-0.69

-7.60

-5.30

Correct answer:

-7.60

Explanation:

We're told that the rate of change 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 from 20,000 to 10 between 4 AM and 5 AM, we can solve for this constant of proportionality:

$10=20000e^{k(5-4)}$

$\frac{1}{2000}=e^{k}$

$k=ln(\frac{1}{2000})=-7.60$

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Example Question #145 : Constant Of Proportionality

The rate of change of the number of skeletons in the skeleton army is proportional to the population. The population increased by 66.6 percent between September and October. What is the constant of proportionality in months^-1?

Possible Answers:

0.510

0.265

0.971

0.030

1.045

Correct answer:

0.510

Explanation:

We're told that the rate of change 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 increased by 66.6 percent between September and October, we can solve for this constant of proportionality (write the months as their number in the calendar):

$(1+0.666)y_0=y_0e^{k(10-9)}$

$1.666=e^{k}$

k=ln(1.666)=0.510

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Example Question #146 : Constant Of Proportionality

The rate of change of the vampires is proportional to the population. The population increased by 11.9 percent between August and October. What is the constant of proportionality in months^-1?

Possible Answers:

0.531

0.335

0.056

0.155

0.815

Correct answer:

0.056

Explanation:

We're told that the rate of change 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 increased by 11.9 percent between August and October, we can solve for this constant of proportionality (use the months' numbers in the calendar):

$(1+0.119)y_0=y_0e^{k(10-8)}$

$1.119=e^{2k}$

2k=ln(1.119)

$k=\frac{ln(1.119)}{2}=0.056$

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Example Question #147 : Constant Of Proportionality

The rate of change of the mummies is proportional to the population. The population increased by 123.4 percent between July and October. What is the constant of proportionality in months^-1?

Possible Answers:

0.449

0.047

0.012

0.113

0.268

Correct answer:

0.268

Explanation:

We're told that the rate of change 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 increased by 123.4 percent between July and October, we can solve for this constant of proportionality (write the months by their number in the calendar):

$(1+1.234)y_0=y_0e^{k(10-7)}$

$2.234=e^{3k}$

3k=ln(2.234)

$k=\frac{ln(2.234)}{3}=0.268$

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Example Question #148 : Constant Of Proportionality

The rate of change of the number of werewolves is proportional to the population. The population increased by 23.4 percent between September and October. What is the constant of proportionality in months^-1?

Possible Answers:

0.089

0.210

0.313

0.101

0.420

Correct answer:

0.210

Explanation:

We're told that the rate of change 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 increased by 23.4 percent between September and October, we can solve for this constant of proportionality (write the months as their number in the calendar):

$(1+0.234)y_0=y_0e^{k(10-9)}$

$1.234=e^{k}$

k=ln(1.234)=0.210

Report an Error

Example Question #3866 : Calculus

The rate of change of the number of zombies is proportional to the population. The population increased by 1523 percent between October 1st and October 31st. What is the constant of proportionality in days^-1?

Possible Answers:

0.434

0.578

0.093

0.378

0.188

Correct answer:

0.093

Explanation:

We're told that the rate of change 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 increased by 1523 percent between October 1st and October 31st, we can solve for this constant of proportionality:

$(1+15.23)y_0=y_0e^{k(31-1)}$

$16.23=e^{30k}$

tk=ln()

$k=\frac{ln(16.23)}{30}=0.093$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #141 : Constant Of Proportionality

Example Question #142 : Constant Of Proportionality

Example Question #143 : Constant Of Proportionality

Example Question #144 : Constant Of Proportionality

Example Question #3861 : Calculus

Example Question #145 : Constant Of Proportionality

Example Question #146 : Constant Of Proportionality

Example Question #147 : Constant Of Proportionality

Example Question #148 : Constant Of Proportionality

Example Question #3866 : Calculus

All Calculus 1 Resources

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