Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 4

Differentiation

Applications of the Derivative

In this unit, we apply the derivative to estimate values of quantities and errors in measurement. Then, we see how the derivative is used in solving problems in the natural and social sciences.

The capital Greek letter delta, $\text{Δ}$, together with a variable, is used to denote a small quantity of the indicated variable. So, $\text{Δ}x$ indicates a small quantity of the variable $x$ (positive or negative); similarly, $\text{Δ}y$ is a small quantity of the variable $y$.

When we write $a+\text{Δ}x$, we indicate the number $a$ plus a small quantity of the variable $x$, and this quantity is on the $x$-axis. The quantity $b+\text{Δ}y$ is on the $y$-axis. The quantity $a+\text{Δ}x$ is close to $a$, since $\text{Δ}x$ is small. It is close to $a$ from the right if $\text{Δ}x>0$, and from the left if $\text{Δ}x$ is negative.

We have pointed out that the tangent line is the line closest to the curve, and that we use it to approximate values of the function at a point close to $f\left(a\right)$. We will go over this concept again, using the delta notation.

First, observe that since $a+\text{Δ}x$ is a quantity, we can evaluate a function at this number; hence, we find $f\left(a+\text{Δ}x\right)$ by replacing $x$ by $a+\text{Δ}x$ in the function $f\left(x\right)$.

For instance, if $f\left(x\right)=4x^2+\sqrt{x}$, then

$\begin{array}{ll}\hfill f\left(a+\text{Δ}x\right)& =4{\left(a+\text{Δ}x\right)}^2+\sqrt{a+\text{Δ}x}\hfill \\ \hfill & =4\left(a^2+2a\text{Δ}x+{\left(\text{Δ}x\right)}^2\right)+\sqrt{a+\text{Δ}x}\text{.}\hfill \end{array}$

Remember that $\text{Δ}x$ is a number.

Figure 4.18.$a+\text{Δ}x$ close to $a$ from the right

In Figure 4.18, above, we have $a+\text{Δ}x$ close to $a$ from the right; that is, $\text{Δ}x>0$.

We see that $f\left(a+\text{Δ}x\right)-f\left(a\right)$ is a small quantity on the $y$-axis; hence, we write

$\text{Δ}y=f\left(a+\text{Δ}x\right)-f\left(a\right)\text{.}$

The equation of the tangent line at the point $\left(a,f\left(a\right)\right)$ is

$L\left(x\right)=f^{\prime }\left(a\right)\left(x-a\right)+f\left(a\right)\text{,}$

as we indicated in Proposition 4.6.

Then $L\left(a+\text{Δ}x\right)=f^{\prime }\left(a\right)\left(\left(a+\text{Δ}x\right)-a\right)+f\left(a\right)$, and

$L\left(a+\text{Δ}x\right)=f^{\prime }\left(a\right)\left(\text{Δ}x\right)+f\left(a\right)$

and

$f\left(a+\text{Δ}x\right)\approx L\left(a+\text{Δ}x\right)=f^{\prime }\left(a\right)\left(\text{Δ}x\right)+f\left(a\right)\text{.}$ (4.4)

Hence,

$\text{Δ}y=f\left(a+\text{Δ}x\right)-f\left(a\right)\approx L\left(a+\text{Δ}x\right)-f\left(a\right)=f^{\prime }\left(a\right)\left(\text{Δ}x\right)\text{.}$ (4.5)

From Figure 4.18, we see that these two approximations are indeed true. There are three important consequences of Equations (4.4) and (4.5).

1. From (4.4), we obtain the linear approximation of $f\left(a+\text{Δ}x\right)$; that is

   $f\left(a+\text{Δ}x\right)\approx f^{\prime }\left(a\right)\left(\text{Δ}x\right)+f\left(a\right)\text{.}$

   We used this approximation in Examples 4.20 and 4.21. The equation of the tangent line $y=f^{\prime }\left(a\right)\left(\text{Δ}x\right)+f\left(a\right)$ is called the linearization of $f$ at $a\text{.}$

2. We also find that

   ${\displaystyle \text{Δ}y=f\left(a+\text{Δ}x\right)-f\left(a\right)\approx f^{\prime }\left(a\right)\left(\text{Δ}x\right)\text{.}}$

3. From (4.5), we have ${\displaystyle {\displaystyle \frac{\text{Δ}y}{\text{Δ}x}}\approx f^{\prime }\left(a\right)\text{.}}$ Then

   ${\displaystyle f^{\prime }\left(a\right)\approx {\displaystyle \frac{\text{Δ}y}{\text{Δ}x}}={\displaystyle \frac{f\left(a+\text{Δ}x\right)-f\left(a\right)}{\text{Δ}x}}\text{.}}$

The next examples show the applications of A, B and C, respectively.

When we know the exact value of $f\left(a\right)$, but not of $f\left(a+\text{Δ}x\right)$, we estimate the latter value using linear approximation.

Example 4.52. We have

${\displaystyle cos \left(30\text{°}\right)=cos \left({\displaystyle \frac{\pi }{6}}\right)={\displaystyle \frac{\sqrt{3}}{2}}}$

but we cannot find an exact value for

${\displaystyle cos \left(32\text{°}\right)=cos \left({\displaystyle \frac{8\pi }{45}}\right)\text{.}}$

So we use linear approximation to estimate $cos \left(32\text{°}\right)$.

Since

${\displaystyle {\displaystyle \frac{8\pi }{45}}={\displaystyle \frac{\pi }{6}}+{\displaystyle \frac{\pi }{90}}\text{,}}$

we can say that

${\displaystyle a={\displaystyle \frac{\pi }{6}}\text{ and }\text{Δ}x={\displaystyle \frac{\pi }{90}}\text{.}}$

From consequence A, above, we have

${\displaystyle cos \left({\displaystyle \frac{8\pi }{45}}\right)=cos \left({\displaystyle \frac{\pi }{6}}+{\displaystyle \frac{\pi }{90}}\right)\approx \left[{\displaystyle \frac{d}{dx}}cos x{|}_{x=\pi ∕6}\right]\left({\displaystyle \frac{\pi }{90}}\right)+cos \left({\displaystyle \frac{\pi }{6}}\right)\text{.}}$

Finding the derivative gives us

${\displaystyle {\displaystyle \frac{d}{dx}} cos x{|}_{x=\pi ∕6}=-sin x{|}_{x=\pi ∕6}=-sin \left({\displaystyle \frac{\pi }{6}}\right)=-{\displaystyle \frac{1}{2}}\text{.}}$

Hence,

${\displaystyle cos \left(32\text{°}\right)=cos \left({\displaystyle \frac{8\pi }{45}}\right)\approx -{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{\pi }{90}}\right)+{\displaystyle \frac{\sqrt{3}}{2}}=-{\displaystyle \frac{\pi }{180}}+{\displaystyle \frac{\sqrt{3}}{2}}={\displaystyle \frac{90\sqrt{3}-\pi }{180}}\text{.}}$

This example is shown graphically in Figure 4.19, below.

Figure 4.19. Graph showing the tangent line at $8\pi ∕45$ compared with $cos 32$°

Since ${\displaystyle y\left({\displaystyle \frac{8\pi }{45}}\right)}$ is above the graph of $cos x$, we can see that our approximation is an overestimate of the actual value of $cos \left(32\text{°}\right)$. That is

${\displaystyle y\left({\displaystyle \frac{8\pi }{45}}\right)={\displaystyle \frac{90\sqrt{3}-\pi }{180}}>cos \left(32\text{°}\right)\text{.}}$

Example 4.53. To estimate the value of ${\left(0\text{.}89\right)}^{4∕3}\text{,}$ we first recognize that we are looking for $f\left(0\text{.}89\right)\text{,}$ which is equal to ${\left(0\text{.}89\right)}^{4∕3}\text{.}$ So the function is $f\left(x\right)=x^{4∕3}\text{.}$

Next, we look for a value $a$ close to $0\text{.}89$ such that $f\left(a\right)$ is known. We see that $0\text{.}89=1-0\text{.}11\text{,}$ and we know that $f\left(1\right)=1\text{.}$ So we let $a=1$ and $\text{Δ}x=-0\text{.}11\text{.}$ Then from conclusion A, above, we recognize that

${\left(0\text{.}89\right)}^{4∕3}=f\left(0\text{.}89\right)\approx f^{\prime }\left(1\right)\left(\text{Δ}x\right)+f\left(1\right)\text{.}$

Since ${\displaystyle f^{\prime }\left(x\right)=\left({\displaystyle \frac{4}{3}}\right)\left(x^{1∕3}\right)\text{,}}$ we find that ${\displaystyle f^{\prime }\left(1\right)={\displaystyle \frac{4}{3}}\text{.}}$

Hence,

${\displaystyle {\left(0\text{.}89\right)}^{4∕3}\approx \left({\displaystyle \frac{4}{3}}\right)\left(-0\text{.}11\right)+1={\displaystyle \frac{64}{75}}\text{.}}$

The application of conclusion B, above, is used in estimating errors in measurements. When an error in measurement occurs, it affects further calculations, a process called error propagation.

From conclusion B, we know that the propagated error $\text{Δ}y$ is approximately equal to $f^{\prime }\left(a\right)\left(\text{Δ}x\right)$. This expression has a special name.

Hence, by conclusion B,

$\text{Δ}y\approx dy\text{;}\text{ that is, }f\left(a+\text{Δ}x\right)-f\left(a\right)\approx f^{\prime }\left(a\right)\text{Δ}x\text{.}$

The relative error can be estimated using differentials:

${\displaystyle {\displaystyle \frac{\text{Δ}y}{f\left(a\right)}}\approx {\displaystyle \frac{dy}{f\left(a\right)}}\text{.}}$

It is always easier to calculate the differential $dy$ than the difference $\text{Δ}y$.

Example 4.54. A metal rod $20$ cm long and $6$ cm in diameter is to be covered (except for the ends) with insulation that is $0\text{.}2$ cm thick. We want to estimate the amount of insulation needed. The exact amount of insulation is equal to the difference of the volume of the rod plus the insulation (radius $3\text{.}2$ cm) and the volume of the rod without the insulation (radius $3$ cm).

Let $V$ be the volume of the rod, and let $r$ be its radius. So, $V=20\pi r^2\text{,}$ and the amount of insulation is

$V\left(3\text{.}2\right)-V\left(3\right)=\text{Δ}V=20\pi {\left(3\text{.}2\right)}^2-20\pi 3^2=20\pi \left({\left(3\text{.}2\right)}^2-3^2\right)\text{.}$

We could evaluate this amount, or use differentials to estimate its value, as follows:

$\text{Δ}V=V\left(3\text{.}2\right)-V\left(3\right)\approx dV=V^{\prime }\left(3\right)\left(0\text{.}2\right)\text{.}\text{[Why?]}$

We determine that $V^{\prime }\left(r\right)=40\pi r$, then

$\text{Δ}V\approx 40\pi \left(3\right)\left(0\text{.}2\right)=24\pi {\text{ cm}}^3\text{.}$

You can check that ${\displaystyle \text{Δ}V={\displaystyle \frac{124\pi }{5}}}\text{,}$ and ${\displaystyle {\displaystyle \frac{124\pi }{5}}\approx 24\pi \text{.}}$

Note the advantage that it was easier to calculate $dV$ than $\text{Δ}V$.

The relative error in this case is approximately equal to

${\displaystyle {\displaystyle \frac{dV}{V\left(3\right)}}={\displaystyle \frac{24\pi }{180\pi }}={\displaystyle \frac{2}{15}}\text{,}}$

and the percentage error is estimated at $13\text{.}33\%$.

Example 4.55. The electrical resistance $R$ of a wire is given by

${\displaystyle R\left(r\right)={\displaystyle \frac{k}{r^2}}\text{,}}$

where $k$ is a constant and $r$ is the radius of the wire. If the percentage error of the radius must be $\pm 4\%\text{,}$ what is the percentage error of $R$?

The relative error of ${\displaystyle {\displaystyle \frac{\text{Δ}R}{R}}}$ is approximately equal to ${\displaystyle {\displaystyle \frac{dR}{R}}\text{.}}$ Hence,

${\displaystyle {\displaystyle \frac{\text{Δ}R}{R}}\approx {\displaystyle \frac{dR}{R}}={\displaystyle \frac{R^{\prime }\left(r\right)\left(\text{Δ}R\right)}{R}}={\displaystyle \frac{\left(-2k{r}^{-3}\right)\left(\text{Δ}r\right)}{k{r}^{-2}}}=\left(-2\right){\displaystyle \frac{\text{Δ}r}{r}}\text{.}}$

Since the relative error ${\displaystyle {\displaystyle \frac{\text{Δ}r}{r}}}$ must be between $-0\text{.}04$ and $0\text{.}04\text{,}$ we have

${\displaystyle -0\text{.}04\le {\displaystyle \frac{\text{Δ}r}{r}}\le 0\text{.}04\text{.}}$

Multiplying by $-2$ yields

${\displaystyle -0\text{.}04\left(-2\right)=0\text{.}08\ge \left(-2\right){\displaystyle \frac{\text{Δ}r}{r}}\ge 0\text{.}04\left(-2\right)=-0\text{.}08\text{.}}$

We find that the relative error of $R$ is approximately $\pm 8$%.

Going back to the interpretation of the derivative as the instantaneous rate of change, we know, by conclusion C, that

${\displaystyle f^{\prime }\left(a\right)\approx {\displaystyle \frac{\text{Δ}y}{\text{Δ}x}}={\displaystyle \frac{f\left(a+\text{Δ}x\right)-f\left(a\right)}{\text{Δ}x}}\text{.}}$

Observe that the quotient

${\displaystyle {\displaystyle \frac{f\left(a+\text{Δ}x\right)-f\left(a\right)}{\text{Δ}x}}}$

is the average rate of change. So, the instantaneous rate of change (derivative) is its limit:

${\displaystyle f^{\prime }\left(a\right)=\underset{\text{Δ}x\to 0}{\lim }{\displaystyle \frac{f\left(a+\text{Δ}x\right)-f\left(a\right)}{\text{Δ}x}}=\underset{\text{Δ}x\to 0}{\lim }{\displaystyle \frac{\text{Δ}y}{\text{Δ}x}}\text{.}}$

We look into the average rate of change first, and then we find the instantaneous rate of change. Note that in applications in the natural and social sciences—physics, chemistry, biology and economics—the concepts of velocity, linear density, rate of reaction, compressibility, rate of growth, marginal cost and marginal revenue are all instantaneous rates of change.

Example 4.56. The linear density $\rho$ of a nonhomogeneous rod is the limit of the average density of the rod. If the mass measured from its left end point is $x$, then the mass is $m=f\left(x\right)\text{,}$ and the average density is defined as mass over the length. Hence, the average density is

${\displaystyle {\displaystyle \frac{\text{Δ}m}{\text{Δ}x}}={\displaystyle \frac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}}\text{,}}$

and the linear density is

${\displaystyle \rho =\underset{\text{Δ}x\to 0}{\lim }{\displaystyle \frac{\text{Δ}m}{\text{Δ}x}}=f^{\prime }\left(x\right)\text{.}}$

If the mass is $f\left(x\right)=3x^2$ kg, then the linear density is $f^{\prime }\left(x\right)=6x$ kg/m. The linear density at $x=1$ m is $f^{\prime }\left(1\right)=6$ kg/m; at $x=2$ m, it is $f^{\prime }\left(2\right)=12$ kg/m; and at $x=3$ m, it is $f^{\prime }\left(3\right)=18$ kg/m. Thus, the linear density increases, and it will be at its highest at the end of the rod. See Exercise 17 on page 178 of the textbook.

Example 4.57. See Exercise 28 on page 180 of the textbook.

Let $f$ be the frequency of vibrations of a vibrating violin string

${\displaystyle f={\displaystyle \frac{1}{2L}}\sqrt{{\displaystyle \frac{T}{\rho }}}}$

where $L$ is the length of the string, $T$ is the tension, and $\rho$ is the linear density.

1. The rate of change of the frequency with respect to
   1. the length ($T$ and $\rho$ are constant) is

      ${\displaystyle f^{\prime }\left(L\right)={\displaystyle \frac{df}{dL}}=\left({\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{T}{\rho }}}\right)\left({\displaystyle \frac{d}{dL}}{\displaystyle \frac{1}{L}}\right)=-{\displaystyle \frac{1}{2L^2}}\sqrt{{\displaystyle \frac{T}{\rho }}}\text{.}}$

   2. the tension ($L$ and $\rho$ are constant) is

      ${\displaystyle f^{\prime }\left(T\right)={\displaystyle \frac{df}{dT}}=\left({\displaystyle \frac{1}{2L}}{\displaystyle \frac{1}{\sqrt{\rho }}}\right)\left({\displaystyle \frac{d}{dT}}\sqrt{T}\right)={\displaystyle \frac{1}{4L\sqrt{T}\rho }}\text{.}}$

   3. the linear density ($L$ and $T$ are constant) is

      ${\displaystyle f^{\prime }\left(\rho \right)={\displaystyle \frac{df}{d\rho }}=\left({\displaystyle \frac{1}{2L^2}}\sqrt{T}\right)\left({\displaystyle \frac{d}{d\rho }}{\displaystyle \frac{1}{\sqrt{\rho }}}\right)=-{\displaystyle \frac{\sqrt{T}}{4L{\rho }^{3∕2}}}\text{.}}$

2. What happens to the pitch of the note when
   1. the effective length of the string is decreased by placing a finger on the string so that a shorter portion of the string vibrates?

      The frequency is of the form $f\left(L\right)=k∕L$ for $k$ a constant (sketch a possible graph of this function). Since the derivative $f^{\prime }\left(L\right)$ is negative, the frequency decreases for any increase in $L$, giving us a lower note. Conversely, when $L$ decreases, the frequency increases, so we have a higher note.

   2. the tension is increased by turning a tuning peg?

      The frequency is of the form $f\left(T\right)=k\sqrt{T}$ for $k$ a constant (sketch a possible graph of this function). The derivative is positive, so the frequency increases and when $T$ increases, giving us a higher note.

   3. the linear density is increased by switching to another string?

      The frequency is of the form ${\displaystyle f\left(\rho \right)=k{\displaystyle \frac{1}{\sqrt{\rho }}}}$ for $k$ a constant (sketch a possible graph of this function). The derivative is negative, so the frequency decreases and when $\rho$ increases, giving a lower note.