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Calculus 1 : How to find constant of proportionality of rate

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Example Question #181 : Constant Of Proportionality

The rate of change of the number of constant of proportionality calculus problems is proportional to the population. The population increased from 15 to 6750 between September and October. What is the constant of proportionality in months^-1?

Possible Answers:

3.23

1.92

6.11

4.07

5.32

Correct answer:

6.11

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 15 to 6750 between September and October, we can solve for this constant of proportionality (it is useful to treat the months as their number in the calendar):

$6750=15e^{k(10-9)}$

$450=e^{k}$

k=ln(450)=6.11

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Example Question #182 : Constant Of Proportionality

What is the constant of proportionality of s(t)=6ln(4t) between t=3 and t=6 ?

Possible Answers:

c=2ln(24)-2ln(12)

c=6ln(6)

c=6ln(12)

c=6ln(24)

Correct answer:

c=2ln(24)-2ln(12)

Explanation:

The constant of proportionality between t=a and t=b is given by the equation

$c=\frac{\Delta y}{\Delta x}=\frac{s(b)-s(a)}{b-a}$

In this problem,

s(3)=6ln(4*3)=6ln(12)

s(6)=6ln(4*6)=6ln(24)

$c=\frac{s(b)-s(a)}{b-a}=\frac{6ln(24)-6ln(12)}{6-3}=\frac{6ln(24)-6ln(12)}{3}$

c=2ln(24)-2ln(12)

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Example Question #183 : Constant Of Proportionality

132N of force is required to stretch a spring 12m . What is the constant of proportionality of the spring?

Possible Answers:

F=11N/m

F=12N/m

F=132N/m

F=120N/m

Correct answer:

F=11N/m

Explanation:

The relation between the force and stretch of a spring is

F=kx where is force, is the spring constant, or proportionality of the spring, and is how far the spring is strectched.

For this problem

$F=kx\Rightarrow k=\frac{F}{x}=\frac{132}{12}=11N/m$

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Example Question #184 : Constant Of Proportionality

What is the constant of proportionality of a circle with a diameter of 3in and a circumference of 9.42in ?

Possible Answers:

9.42

$\pi$

Correct answer:

$\pi$

Explanation:

The relation between circumference and diameter is

$C=\pi D$ where is the circumference of a circle and is the diameter of the circle.

The constant of proportionality is $p=C/D= \pi$ for all circles.

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Example Question #185 : Constant Of Proportionality

The population of a town grows exponentially from 200K to 800K in 3yrs . What is the population growth constant?

Possible Answers:

$k=\frac{ln(4)}{3}$

k=3

k=4

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

Correct answer:

$k=\frac{ln(4)}{3}$

Explanation:

Exponential growth is modeled by the equation $A=Pe^{kt}$

where is the final amount, is the inital amount, is the growth constant and is time.

In this problem, A=800K , P=200K and t=3 . Substituting these variables into the growth equation the solving for gives us

$800,000=200,000e^{3k}$

$\frac{800,000}{200,000}=e^{3k}$

$4=e^{3k}$

$ln(4)=ln(e^{3k})$

ln(4)=3k

$k=\frac{ln(4)}{3}$

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Example Question #186 : Constant Of Proportionality

Cobalt-60 has a half-life of 5.27yrs . What is the decay constant of Cobalt-60?

Possible Answers:

k=-0.5

$k=\frac{ln(0.5)}{60}$

$k=\frac{ln(60)}{5.27}$

$k=\frac{ln(0.5)}{5.27}$

Correct answer:

$k=\frac{ln(0.5)}{5.27}$

Explanation:

The half-life of an isotope is the time it takes for half the isotope to disappear. Isotopes decay exponentially.

Exponential decay is also modeled by the equation $A=Pe^{kt}$

where is the final amount, is the inital amount, is the growth constant and is time.

Since half the isotope has disappeared, the final amount is half the inital amount , or $\frac{A}{P}=0.5$ .

In this problem, t=5.27yrs .

Substituting these variables into the exponential equation and solving for gives us

$A=Pe^{kt}\Rightarrow \frac{A}{P}=e^{kt}$

$0.5=e^{5.27k}$

$ln(0.5)=ln(e^{5.27k})$

ln(0.5)=5.27k

$k=\frac{ln(0.5)}{5.27}$

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Example Question #187 : Constant Of Proportionality

The number of cats double every 64dys . How many cats will there be after 1 yr if there are cats initially?

Possible Answers:

$A\approx 730$

$A\approx 64$

$A\approx 104$

$A\approx 365$

Correct answer:

$A\approx 104$

Explanation:

Exponential growth is modeled by the equation $A=Pe^{kt}$

where is the final amount, is the inital amount, is the growth constant and is time.

After 64dys , the number of cats has doubled, or the final amount is double the inital amount , or $\frac{A}{P}=2$ .

In this problem, t=64 .

Substituting these variables into the exponential equation and solving for gives us

$A=Pe^{kt}\Rightarrow \frac{A}{P}=e^{kt}$

$2=e^{64k}$

$ln(2)=ln(e^{64k})$

ln(2)=64k

$k=\frac{ln(2)}{64}$

To find the number cats after year, P=2 , $k=\frac{ln(2)}{64}$ , and t=365

$A=Pe^{kt}=2e^{\frac{ln(2)}{64}*365}\approx 104$

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Example Question #188 : Constant Of Proportionality

The number of students enrolled in college has increased by $37\%$ every year since 2003 . If 460K students enrolled in 2003 , how many student enrolled in 2013 ?

Possible Answers:

y=10.3K

y=460M

y=2

y=10.7M

Correct answer:

y=10.7M

Explanation:

The exponential growth is modeled by the equation $y=a(1+r)^{t}$

where is the final amount, is the inital amount, is the growth rate and is time.

In this problem, a=460K , r=0.37 , and t=2013-2003=10 . Substituting these values into the equation gives us

$y=460,000(1+0.37)^{10}=10.7M$

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Example Question #189 : Constant Of Proportionality

The number of CD players owned has decreased by $15\%$ annually since 2008 . If people owned CD players in 2008 , how many people owned CD players in 2014 ?

Possible Answers:

y=1.13M

y=3K

y=390K

y=11M

Correct answer:

y=1.13M

Explanation:

The exponential growth is modeled by the equation $y=a(1+r)^{t}$

where is the final amount, is the inital amount, is the growth rate and is time.

In this problem, a=3M and t=2014-2008=6 . r=-0.15 because the rate is decreasing. Substituting these values into the equation gives us

$y=3,000,000(1-0.15)^{6}=1.13M$

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Example Question #190 : Constant Of Proportionality

You deposit $\$1500$ into your savings account. After 8 yrs , your account has $\$6500$ in it. What is the interest rate of this account if the account was untouched during the 8 yrs ?

Possible Answers:

$r=20.12\%$

$r=2.38\%$

$r=40.76\%$

$r=0.85\%$

Correct answer:

$r=20.12\%$

Explanation:

The exponential growth is modeled by the equation $y=a(1+r)^{t}$

where is the final amount, is the inital amount, is the growth rate and is time.

In this problem, $a=\$1500$ , $y=\$6500$ and t=8 yrs . Substituting these variables into the growth equation and solving for r gives us

$6500=1500(1+r)^{8}$

$\frac{13}{3}=(1+r)^{8}$

$\sqrt[8]{\frac{13}{3}}=\sqrt[8]{(1+r)^{8}}$

$\sqrt[8]{\frac{13}{3}}=1+r$

1+r=1.2012

$r=0.2012=20.12\%$

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Calculus 1 : How to find constant of proportionality of rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #181 : Constant Of Proportionality

Example Question #182 : Constant Of Proportionality

Example Question #183 : Constant Of Proportionality

Example Question #184 : Constant Of Proportionality

Example Question #185 : Constant Of Proportionality

Example Question #186 : Constant Of Proportionality

Example Question #187 : Constant Of Proportionality

Example Question #188 : Constant Of Proportionality

Example Question #189 : Constant Of Proportionality

Example Question #190 : Constant Of Proportionality

All Calculus 1 Resources

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