OBJECTIVES

When you finish this section, you should be able to:

1. Find the differential of a function and interpret it geometrically (p. 230)
2. Find the linear approximation to a function (p. 232)
3. Use differentials in applications (p. 233)
4. Use Newton’s Method to approximate a real zero of a function (p. 234)

DEFINITION

Let $y=f(x)$ be a differentiable function and let $\Delta x$ denote a change in $x$.

The differential of $x$, denoted $dx$, is defined as $dx=\Delta x≠ 0$.

The differential of $y$, denoted $dy$, is defined as $$\bbox[5px, border:1px solid black, #F9F7ED]{ {dy=f^\prime (x) dx} }\tag{2}$$

For the function $f(x) =xe^{x}$:

Solution (a) $dy=f^\prime (x) dx= (xe^{x}+e^{x}) dx=(x+1) e^{x}dx$.

(b) See Figure 9(a). When $x=0$ and $\Delta x=dx=0.5$, then $$\ dy=(x+1) e^{x}dx=(0+1) e^{0}(0.5) =0.5$$

Figure 9

The tangent line rises by $0.5$ as $x$ changes from $0$ to $0.5$. The corresponding change in the height of the graph $f$ is $$\begin{eqnarray*} \Delta y=f( x+\Delta x) -f( x) &=&f(0.5) -f( 0) =0.5e^{0.5}-0\approx 0.824\\ \vert \Delta y-dy\vert &\approx& \vert 0.824-0.5\vert =0.324 \end{eqnarray*}$$

The graph of the tangent line is approximately $0.324$ below the graph of $f$ at $x\,{=}\,0.5$.

(c) See Figure 9(b). When $x=0$ and $\Delta x=dx=0.1$, then $$dy=(0+1) e^{0}(0.1) =0.1$$

The tangent line rises by $0.1$ as $x$ changes from $0$ to $0.1.$ The corresponding change in the height of the graph $f$ is $$\begin{eqnarray*} \Delta y=f( x+\Delta x) -f( x) &=&f(0.1) -f( 0) =0.1e^{0.1}-0\approx 0.111\\ \vert \Delta y-dy\vert &\approx& \left\vert 0.111-0.1\right\vert =0.011 \end{eqnarray*}$$

The graph of the tangent line is approximately $0.011$ below the graph of $f$ at $x=0.1$.

(d) See Figure 9(c). When $x=0$ and $\Delta x=dx=0.01$, then $$dy=(0+1) e^{0}( 0.01) =0.01$$

The tangent line rises by $0.01$ as $x$ changes from $0$ to $0.01.$ The corresponding change in the height of the graph of $f$ is $$\begin{eqnarray*} \Delta y=f( x+\Delta x) -f( x) &=&f( 0.01) -f( 0) =0.01e^{0.01}-0\approx 0.0101\\ \vert \Delta y-dy\vert &\approx& \vert 0.0101-0.01\vert =0.0001 \end{eqnarray*}$$

The graph of the tangent line is approximately $0.0001$ below the graph of $f$ at $x=0.01$.

(e) The closer $\Delta x$ is to $0$, the closer $dy$ is to $\Delta y.$ So, we conclude that the closer $\Delta x$ is to $0$, the less the tangent line departs from the graph of the function. That is, we can use the tangent line to $f$ at a point $P$ as a linear approximation to $f$ near $P$.

NOW WORK

Problems 9 and 17.

THEOREM Linear Approximation

The linear approximation ${L( x)}$ to a differentiable function $f$ at $x=x_{0}$ is given by $$\bbox[5px, border:1px solid black, #F9F7ED]{L( x) =f( x_{0}) +f^\prime ( x_{0}) ( x-x_{0})} \tag{3}$$

Solution (a) Since $f^\prime ( x) =\cos x$, then $f(0)=\sin 0=0$ and $f^\prime (0) = \cos 0 = 1$. Using Equation (3), the linear approximation $L( x)$ to $f$ at $0$ is $$L( x) =f(0) + f^\prime (0)(x -0)=x$$

So, for $x$ close to $0,$ the function $f( x) =\sin x$ can be approximated by the line $L( x) =x.$

(b) The approximate values of $\sin x$ using $L( x) =x,$ the true values of $\sin x,$ and the absolute error in using the approximation are given in Table 3. From Table 3, we see that the further $x$ is from $0,$ the worse the line $L( x) =x$ approximates $f( x) =\sin x$.

(c) See Figure 11 for the graphs of $f( x) =\sin x$ and $L( x) =x$.

NOW WORK

Problem 25.

A spherical bearing has a radius of $3{\,{\rm{cm}}}$ when it is new. Use differentials to approximate the volume of the metal lost after the bearing wears down to a radius of $2.971{\,{\rm{cm}}}$. Compare the approximation to the actual volume lost.

Solution The volume $V$ of a sphere of radius $R$ is $V=\dfrac{4}{3}\pi R^{3}.$ As a machine operates, friction wears away part of the bearing. The exact volume of metal lost equals the change $\Delta V$ in the volume $V$ of the sphere, when the change in the radius of the bearing is $\Delta R=2.971-3={-}0.029{\,{\rm{cm}}}$. Since the change $\Delta R$ is small, we use the differential $dV$ to approximate the change $\Delta V$. Then $$\begin{eqnarray*} && \Delta V \approx dV \underset{\underset{\color{#0066A7}{\hbox{\(dV=V^{\,\prime} dR\)}}}{\color{#0066A7}{\uparrow }}}{=} 4\pi R^{2}dR \underset{\underset{\color{#0066A7}{\hbox{\(dR=\Delta R\)}}}{\color{#0066A7}{\uparrow}}}{=} (4\pi )(3^{2})({-} 0.029) \approx {-}3.280 \end{eqnarray*}$$

The approximate loss in volume of the bearing is $3.28{\,{\rm{cm}}}^{3}$.

The actual loss in volume $\Delta V$ is $$\Delta V=V(R+\Delta R)-V(R)=\dfrac{4}{3}\pi \cdot 2.971^{3}-\dfrac{4}{3}\pi \cdot 3^{3}=\dfrac{4}{3}\pi ( -0.7755) =-3.248{\,{\rm{cm}}}^{3}$$

The approximate change in volume is correct to one decimal place.

NOW WORK

Problem 47.

A company manufactures spherical ball bearings of radius $3{\,{\rm{cm}}}$. The customer accepts a tolerance of 1% in the radius. Use differentials to approximate the relative error for the surface area of the acceptable ball bearings.

Solution The tolerance of 1% in the radius $R$ means that the relative error in the radius $R$ must be within $0.01.$ That is, $\dfrac{\vert \Delta R\vert }{R}≤ 0.01$. The surface area $S$ of a sphere of radius $R$ is given by the formula $S=4\pi R^{2}.$ We seek the relative error in $S, \dfrac{\vert \Delta S\vert }{S},$ which can be approximated by $\dfrac{\vert dS\vert}{S}$. $$\begin{eqnarray*} && \frac{\vert \Delta S\vert }{S}\approx \frac{\vert dS\vert }{S} \underset{\underset{\color{#0066A7}{\hbox{\(dS=S^{\,\prime} dR\)}}}{\color{#0066A7}{\uparrow}}}{=} \frac{( 8\pi R) \vert dR \vert }{4\pi R^{2}}=\frac{2\vert dR\vert }{R}= 2\cdot \frac{\vert \Delta R\vert }{R} \underset{\underset{\color{#0066A7}{\hbox{\(\tfrac{\vert \Delta R\vert }{R}≤ 0.01\)}}}{\color{#0066A7}{ \uparrow}}}{\le} 2(0.01)= 0.02 \end{eqnarray*}$$

The relative error in the surface area will be less than or equal to $0.02$.

NOW WORK

Problem 59.

Newton’s Method

Suppose a function $f$ is defined on a closed interval $[a, b]$ and its first derivative $f^\prime$ is continuous and not equal to $0$ on the open interval $( a, b) .$ If $x=c_{1}$ is a sufficiently close first approximation to a real zero of the function $f$ in $( a, b)$, then the formula $$\bbox[5px, border:1px solid black, #F9F7ED]{ c_{2}=c_{1}-\dfrac{f(c_{1})}{f^\prime (c_{1})}}$$ gives a second approximation to the zero. The $n$th approximation to the zero is given by $$\bbox[5px, border:1px solid black, #F9F7ED]{ c_{n}=c_{n-1}-\dfrac{f( c_{n-1}) }{f^\prime (c_{n-1})}}$$

ORIGINS

When Isaac Newton first introduced his method for finding real zeros in 1669, it was very different from what we did here. Newton’s original method (also known as the Newton–Raphson Method) applied only to polynomials, and it did not use calculus! In 1690 Joseph Raphson simplified the method and made it recursive, again without calculus. It was not until 1740 that Thomas Simpson used calculus, giving the method the form we use here.

Use Newton’s Method to find a fourth approximation to the positive real zero of the function $f( x) =x^{3}+x^{2}-x-2$.

Solution Since $f(1) =-1$ and $f(2) =8,$ we know from the Intermediate Value Theorem that $f$ has a zero in the interval $(1, 2)$. Also, since $f$ is a polynomial function, both $f$ and $f^\prime$ are differentiable functions. Now $$f(x) =x^{3}+x^{2}-x-2 \qquad \hbox{and}\qquad f^\prime ( x) =3x^{2}+2x-1$$

We choose $c_{1}=1.5$ as the first approximation and use Newton’s Method. The second approximation to the zero is $$c_{2}=c_{1}-\dfrac{f( c_{1}) }{f^\prime ( c_{1}) }=1.5-\dfrac{f(1.5) }{f^\prime (1.5) }=1.5-\dfrac{2.125}{ 8.75}\approx 1.2571429$$

Use Newton’s Method again, with $c_{2}=1.2571429$. Then $f(1.2571429) \approx 0.3100644$ and $f^\prime (1.2571429)\approx 6.2555106$. The third approximation to the zero is $$c_{3}=c_{2}-\dfrac{f( c_{2}) }{f^\prime ( c_{2}) }=1.2571429-\dfrac{0.3100641}{6.2555102}\approx 1.2075763$$

Figure 13 $f(x) = x^{3} + x^{2} - x - 2$

The fourth approximation to the zero is $$c_{4}=c_{3}-\dfrac{f( c_{3}) }{f^\prime ( c_{3}) } =1.2075763-\dfrac{0.0116012}{5.7898745}\approx 1.2055726$$

NOW WORK

Problem 33.

The graph of the function $f( x) =\sin x+x^{2}-1$ is shown in Figure 14.

Figure 14 $f(x) = \sin x + x^{2}-1$

Solution (a) The function $f( x) =\sin x+x^{2}-1$ is continuous on its domain, all real numbers, so it is continuous on the closed interval $[0,1] .$ Since $f( 0) =-1$ and $f(1) =\sin 1\approx 0.841$ have opposite signs, the Intermediate Value Theorem guarantees that $f$ has a zero in the interval $(0,1) .$

(b) We begin by finding $f^\prime ( x) =\cos x+2x.$ To use Newton’s Method with a graphing utility, we enter $$x-\dfrac{\sin x+x^{2}-1}{\cos x+2x} \qquad {\color{#0066A7}{x-\dfrac{{ f}({x}) }{{ f^\prime }({x})}}}$$

Figure 15

into the $Y= \hbox{editor}$, as shown in Figure 15. We create a table by entering the initial value $0.5$ in the $X$ column. The graphing utility computes $$0.5-\dfrac{\sin 0.5+0.5^{2}-1}{\cos 0.5+2(0.5) }=0.64410789$$

and displays $0.64411$ in column $Y_{1}$ next to $0.5.$ The value $Y_{1}$ is the second approximation $c_{2}$ that we use in the next iteration. That is, we enter $0.64410789$ in the $X$ column of the next row, and the new entry in column $Y_{1}$ is the third approximation $c_{3}.$ We repeat the process until we obtain the desired approximation. The fourth approximation to the zero of $f$ is $0.63673$, as shown in Figure 16.

NOW WORK

Problem 41.