Rate - CLEP Calculus

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Question

Find the average rate of change for over the interval .

Answer

The rate of change of a function is the amount it changes over a given amount of time.

In mathematical terms, this can be written as

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

we plug in our values:

$\frac{5^2+5+1-0^2-0-1}{5-0} = \frac{30}{5}=6$

Note that this is only an average because quadratic functions change at different rates depending on where you are in the function's domain.

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Question

Let $f(x) = ln(e^{x}-sin(\pi -\pi x))$ . Use linear approximation to estimate $f(\frac{3}{2})$ .

Answer

Note that:

Therefore, for values relatively close to 1, we can use the formula for dy (the differential form of the derivative) to estimate f at close values.

$dy= \frac{e^{x}+\pi cos(\pi -\pi x)}{e^{x}-sin(\pi -\pi x)}dx$ From log derivative with chain rule.

Since $\frac{3}{2}$ lies $\frac{1}{2}$ to the right of , $dx = \frac{1}{2}$ and for the estimation, so:

$dy = \frac{e^{3/2}+\pi cos(-\pi )}{e^{3/2}} = 1-\frac{\pi }{e^{3/2}}$

So then:

$f(3/2) \approx1+1-\frac{\pi }{e^{3/2}} = 2-\frac{\pi }{e^{3/2}}$

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Question

Using $(1+x)^{k}\approx1+kx$ , approximate the value of $(0.985)^{50}.$

Answer

First, we need to rearrange the given to match the approximation formula. Therefore, $(0.985)^{50}=(1-0.015)^{50}=1-50(0.015)=0.25.\textup{}$

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Question

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its sides when its sides have length $\frac{7}{30}$ ?

Answer

Begin by writing the equations for a cube's dimensions. Namely its volume in terms of the length of its sides:

The rates of change of the volume can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

$3s^2\frac{ds}{dt}=(\phi)\frac{ds}{dt}$

$\phi=3s^2$

$\phi =3(\frac{7}{30})^2=\frac{49}{300}$

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Question

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its sides when its sides have length $\frac{13}{15}$ ?

Answer

Begin by writing the equations for a cube's dimensions. Namely its volume in terms of the length of its sides:

The rates of change of the volume can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

$3s^2\frac{ds}{dt}=(\phi)\frac{ds}{dt}$

$\phi=3s^2$

$\phi =3(\frac{13}{15})^2=\frac{169}{75}$

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Question

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its sides when its sides have length $\frac{1}{30}$ ?

Answer

Begin by writing the equations for a cube's dimensions. Namely its volume in terms of the length of its sides:

The rates of change of the volume can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

$3s^2\frac{ds}{dt}=(\phi)\frac{ds}{dt}$

$\phi=3s^2$

$\phi =3(\frac{1}{30})^2=\frac{1}{300}$

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Question

Find the average rate of change for over the interval .

Answer

The rate of change of a function is the amount it changes over a given amount of time.

In mathematical terms, this can be written as

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

we plug in our values:

$\frac{5^2+5+1-0^2-0-1}{5-0} = \frac{30}{5}=6$

Note that this is only an average because quadratic functions change at different rates depending on where you are in the function's domain.

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Question

Let $f(x) = ln(e^{x}-sin(\pi -\pi x))$ . Use linear approximation to estimate $f(\frac{3}{2})$ .

Answer

Note that:

Therefore, for values relatively close to 1, we can use the formula for dy (the differential form of the derivative) to estimate f at close values.

$dy= \frac{e^{x}+\pi cos(\pi -\pi x)}{e^{x}-sin(\pi -\pi x)}dx$ From log derivative with chain rule.

Since $\frac{3}{2}$ lies $\frac{1}{2}$ to the right of , $dx = \frac{1}{2}$ and for the estimation, so:

$dy = \frac{e^{3/2}+\pi cos(-\pi )}{e^{3/2}} = 1-\frac{\pi }{e^{3/2}}$

So then:

$f(3/2) \approx1+1-\frac{\pi }{e^{3/2}} = 2-\frac{\pi }{e^{3/2}}$

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Question

Using $(1+x)^{k}\approx1+kx$ , approximate the value of $(0.985)^{50}.$

Answer

First, we need to rearrange the given to match the approximation formula. Therefore, $(0.985)^{50}=(1-0.015)^{50}=1-50(0.015)=0.25.\textup{}$

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Question

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its sides when its sides have length $\frac{7}{30}$ ?

Answer

Begin by writing the equations for a cube's dimensions. Namely its volume in terms of the length of its sides:

The rates of change of the volume can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

$3s^2\frac{ds}{dt}=(\phi)\frac{ds}{dt}$

$\phi=3s^2$

$\phi =3(\frac{7}{30})^2=\frac{49}{300}$

Compare your answer with the correct one above

Question

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its sides when its sides have length $\frac{13}{15}$ ?

Answer

Begin by writing the equations for a cube's dimensions. Namely its volume in terms of the length of its sides:

The rates of change of the volume can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

$3s^2\frac{ds}{dt}=(\phi)\frac{ds}{dt}$

$\phi=3s^2$

$\phi =3(\frac{13}{15})^2=\frac{169}{75}$

Compare your answer with the correct one above

Question

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its sides when its sides have length $\frac{1}{30}$ ?

Answer

Begin by writing the equations for a cube's dimensions. Namely its volume in terms of the length of its sides:

The rates of change of the volume can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

$3s^2\frac{ds}{dt}=(\phi)\frac{ds}{dt}$

$\phi=3s^2$

$\phi =3(\frac{1}{30})^2=\frac{1}{300}$

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Question

The rate of decrease of the population of E. coli in response to a particular antibiotic is proportional to the population. The population decreased from 160,000 to 8,000 between 1:00 and 1:30. Determine the expected population at 1:45.

Answer

We're told that the rate of decrease 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 is an initial population value, represents a measure of elapsed time relative to this population value, and is the constant of proportionality.

Since the population decreased from 160,000 to 8,000 between 1:00 and 1:30, we can solve for this constant of proportionality. Treat the minutes as decimals after the hour by dividing by 60:

$8000=160000e^{k(1.5-1)}$

$\frac{1}{20}=e^{0.5k}$

$0.5k=ln(\frac{1}{20})$

$k=\frac{ln(\frac{1}{20})}{0.5}=-5.991$

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

$P=8000e^{(1.75-1.5)(-5.991)} \approx 1789$

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Question

The rate of growth of the culture of bacteria on a dirty plate is proportional to the population. The population increased from 50 to 200 between 1:15 and 1:30. At what point in time approximately would the population be 700?

Answer

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 is an initial population value, represents a measure of elapsed time relative to this population value, and is the constant of proportionality.

Since the population increased from 50 to 200 between 1:15 and 1:30, we can solve for this constant of proportionality. Treat the minutes as decimals after the hour by dividing by 60:

$200=50e^{k(1.5-1.25)}$

$4=e^{0.25k}$

$k=\frac{ln(4)}{0.25}=5.545$

Now that the constant of proportionality is known, we can use it to estimate our time point:

$700=200e^{5.545(t-1.5)}$

$5.545(t-1.5)=ln(\frac{7}{2})$

$t=1.5+\frac{ln(\frac{7}{2})}{5.545} = 1.726 \rightarrow 1:44$

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Question

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?

Answer

The population size after some time is given by:

where is the initial population.

At the start, there were 30 blood cells.

Substitute this value into the given formula.

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

Substitute this into , and solve for .

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

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

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

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

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Question

The rate of decrease of the robomen population due to the transgalactic organic tai cyber conflict is proportional to the population. The population decreased from 6 billion to 450 million between 2063 and 2081. Determine the expected population in 2100.

Answer

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 is an initial population value, represents a measure of elapsed time relative to this population value, and is the constant of proportionality.

Since the population decreased from 6 billion to 450 million between 2063 and 2081, we can solve for this constant of proportionality:

$450000000=6000000000e^{k(2081-2063)}$

$\frac{3}{40}=e^{18k}$

$18k=ln(\frac{3}{40})$

$k=\frac{ln(\frac{3}{40})}{18}=-0.1439$

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

$P=450000000e^{(2100-2081)(-0.1439)} \approx 29228600$

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Question

Find the direct constant of proportionality of from to .

Answer

To determine the direct constant of proportionality, we determine the rate of change from and for .

Rate of change is determined by

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

In our case, between and , 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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Question

Given any linear function , determine the direct constant of proportionality

Answer

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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Question

Find the direct constant of proportionality of from to .

Answer

Direct constant of proportionality is given by

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

Since and $\Delta x= 4-2=2$

$r= \frac{\Delta y}{\Delta x}=\frac{ln(34)-ln(32)}{2}= \frac{ln(2)}{2}$

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Question

Suppose a population of bacteria increases from to in . What is the constant of growth?

Answer

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 and solve for .

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

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

$1000=100e^{30k}$

$e^{30k}=10$

$30k=\ln(10)$

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

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