# 12.4 Derivatives  (Page 10/18)

 Page 10 / 18

## Real-world applications

For the following exercises, explain the notation in words. The volume $\text{\hspace{0.17em}}f\left(t\right)\text{\hspace{0.17em}}$ of a tank of gasoline, in gallons, $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ minutes after noon.

$f\left(0\right)=600$

$f\text{'}\left(30\right)=-20$

At 12:30 p.m. , the rate of change of the number of gallons in the tank is –20 gallons per minute. That is, the tank is losing 20 gallons per minute.

$f\left(30\right)=0$

$f\text{'}\left(200\right)=30$

At 200 minutes after noon, the volume of gallons in the tank is changing at the rate of 30 gallons per minute.

$f\left(240\right)=500$

For the following exercises, explain the functions in words. The height, $\text{\hspace{0.17em}}s,$ of a projectile after $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ seconds is given by $\text{\hspace{0.17em}}s\left(t\right)=-16{t}^{2}+80t.$

$s\left(2\right)=96$

The height of the projectile after 2 seconds is 96 feet.

$s\text{'}\left(2\right)=16$

$s\left(3\right)=96$

The height of the projectile at $\text{\hspace{0.17em}}t=3\text{\hspace{0.17em}}$ seconds is 96 feet.

$s\text{'}\left(3\right)=-16$

$s\left(0\right)=0,s\left(5\right)=0.$

The height of the projectile is zero at $\text{\hspace{0.17em}}t=0\text{\hspace{0.17em}}$ and again at $\text{\hspace{0.17em}}t=5.\text{\hspace{0.17em}}$ In other words, the projectile starts on the ground and falls to earth again after 5 seconds.

For the following exercises, the volume $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ of a sphere with respect to its radius $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ is given by $\text{\hspace{0.17em}}V=\frac{4}{3}\pi {r}^{3}.$

Find the average rate of change of $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ changes from 1 cm to 2 cm.

Find the instantaneous rate of change of $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ when

$36\pi$

For the following exercises, the revenue generated by selling $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ items is given by $\text{\hspace{0.17em}}R\left(x\right)=2{x}^{2}+10x.$

Find the average change of the revenue function as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ changes from $\text{\hspace{0.17em}}x=10\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}x=20.$

Find $\text{\hspace{0.17em}}R\text{'}\left(10\right)\text{\hspace{0.17em}}$ and interpret.

$50.00 per unit, which is the instantaneous rate of change of revenue when exactly 10 units are sold. Find $\text{\hspace{0.17em}}R\text{'}\left(15\right)\text{\hspace{0.17em}}$ and interpret. Compare $\text{\hspace{0.17em}}R\text{'}\left(15\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}R\text{'}\left(10\right),$ and explain the difference. For the following exercises, the cost of producing $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ cellphones is described by the function $\text{\hspace{0.17em}}C\left(x\right)={x}^{2}-4x+1000.$ Find the average rate of change in the total cost as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ changes from$21 per unit

Find the approximate marginal cost, when 15 cellphones have been produced, of producing the 16 th cellphone.

Find the approximate marginal cost, when 20 cellphones have been produced, of producing the 21 st cellphone.

\$36

## Extension

For the following exercises, use the definition for the derivative at a point $\text{\hspace{0.17em}}x=a,$ $\text{\hspace{0.17em}}\underset{x\to a}{\mathrm{lim}}\frac{f\left(x\right)-f\left(a\right)}{x-a},$ to find the derivative of the functions.

$f\left(x\right)=\frac{1}{{x}^{2}}$

$f\left(x\right)=5{x}^{2}-x+4$

$f\text{'}\left(x\right)=10a-1$

$f\left(x\right)=-{x}^{2}+4x+7$

$f\left(x\right)=\frac{-4}{3-{x}^{2}}$

$\frac{4}{{\left(3-x\right)}^{2}}$

## Finding Limits: A Numerical and Graphical Approach

For the following exercises, use [link] .

$\underset{x\to {-1}^{+}}{\mathrm{lim}}f\left(x\right)$

2

$\underset{x\to {-1}^{-}}{\mathrm{lim}}f\left(x\right)$

$\underset{x\to -1}{\mathrm{lim}}f\left(x\right)$

does not exist

$\underset{x\to 3}{\mathrm{lim}}f\left(x\right)$

At what values of $\text{\hspace{0.17em}}x$ is the function discontinuous? What condition of continuity is violated?

Using [link] , estimate $\text{\hspace{0.17em}}\underset{x\to 0}{\mathrm{lim}}f\left(x\right).$

 $x$ $F\left(x\right)$ −0.1 2.875 −0.01 2.92 −0.001 2.998 0 Undefined 0.001 2.9987 0.01 2.865 0.1 2.78145 0.15 2.678

3

For the following exercises, with the use of a graphing utility, use numerical or graphical evidence to determine the left- and right-hand limits of the function given as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ approaches If the function has limit as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ approaches $\text{\hspace{0.17em}}a,\text{\hspace{0.17em}}$ state it. If not, discuss why there is no limit.

$\begin{array}{ll}\underset{x\to -2}{\mathrm{lim}}\hfill & f\left(x\right)=1\hfill \end{array}$

how to understand calculus?
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
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This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
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hello guys
meena
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It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
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I would like to add that they are used in AC signal analysis for one thing
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Good call Scott. Also radar signals I believe.
Steve
They are used in any profession where the phase of a waveform has to be accounted for in the calculations. Imagine two electrical signals in a wire that are out of phase by 90°. At some times they will interfere constructively, others destructively. Complex numbers simplify those equations
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