12. Differentials & Linear Approximation

a. Differences and Differentials

Differentials and linear approximations are all closely related to tangent lines. So we start by reviewing tangent lines.

Tangent Line

The equation of the tangent line to the graph of the function \(y=f(x)\) at \(x=a\) is: \[ y=f(a)+f'(a)(x-a). \] The right side defines the tangent function: \[ f_{\tan}(x)=f(a)+f'(a)(x-a) \] So the equation of the tangent line is \(y=f_{\tan}(x)\). In differential notation, this is: \[ y=f_{\tan}(x)=f(a)+\left.\dfrac{df}{dx}\right|_{x=a}(x-a). \]

Differences

Given a function \(y=f(x)\), pick a fixed point \(x=a\) and a nearby variable point \(x\).

Differentials

Now construct the tangent line at \(x=a\). While the changes measure values on the graph of the function, the differentials measure values on the graph of the tangent function.

Note that \(dx=\Delta x\). However, \(df\neq \Delta f\), since \(\Delta f\) means the change in a function and \(df\) means the change in the tangent function.

The two formulas for \(dy\) and \(df\) are often written without evaluating the the derivative at \(x=a\). For example: \[ dy=\dfrac{dy}{dx}\,dx \] The evaluation is still understood to be there. However, it is also OK to think we are just cancelling the \(dx\)'s.

If \(x\) is close to \(a\), then \(f_{\tan}(x)\) is a good approximation to \(f(x)\) and \(df\) is a good approximation to \(\Delta f\). This is the linear approximation which is discussed on the next page.

Consider the function \(f(x)=4x^2-x^3\).

1. Find the equation of the tangent line at \(x=2\). (This means \(a=2\).)

   The function, its derivative and their values at \(x=2\) are: \[\begin{aligned} f(x)&=4x^2-x^3 &f(2)&=8 \\ f'(x)&=8x-3x^2 &f'(2)&=4 \end{aligned}\] So the tangent line is \[\begin{aligned} y&=f_{\tan}(x)=f(a)+f'(a)(x-a) \\ &=f(2)+f'(2)(x-2)=8+4(x-2)=4x \end{aligned}\]

2. Find the changes and differentials of \(f\) between \(a=2\) and \(x=2.1\).

   The function value at \(x=2.1\) is: \[ f(2.1)=4\cdot2.1^2-2.1^3=8.379 \] So the change is: \[ \Delta f=f(2.1)-f(2)=8.379-8=.379 \] Remembering that \(f_{\tan}(x)=4x\), the differential is: \[ df=f_{\tan}(2.1)-f(2)=4\cdot2.1-8=.4 \] Notice that \(.4\) is a reasonable approximation to \(.379\).

   We check with an alternate computation of the differential. Remembering that \(f'(2)=4\), the differential is: \[ df=f'(2)dx=4(2.1-2)=.4 \]

1. Find \(V_{\tan}(r)\), the tangent function for the volume at \(r=6\).

   \(V_{\tan}(r)=72\pi+36\pi(r-6)=36\pi\,r-144\pi\)

   The derivative and its value are: \[ V'(r)=\pi r^2 \qquad \qquad V'(6)=\pi 6^2=36\pi \] So the tangent function is \[\begin{aligned} V_{\tan}(r)&=V(6)+V'(6)(r-6) \\ &=72\pi+36\pi(r-6) =36\pi\,r-144\pi \end{aligned}\]

2. If the radius is increased to \(r=h=6.5\,\text{cm}\) (shown in blue), find the change and differential of the volume.

   The change in volume is: \(\Delta V=19.5416\bar{6}\pi\).
   The differential is \(dV=18\pi\).

   The new volume is: \[ V(6.5)=\dfrac{1}{3}\pi 6.5^3=91.5416\bar{6}\pi \] So the change in volume is: \[\begin{aligned} \Delta V &=V(6.5)-V(6) \\ &=91.5416\bar{6}\pi-72\pi=19.5416\bar{6}\pi \end{aligned}\]Remembering that \(V'(6)=36\pi\), the differential is \[\begin{aligned} dV &=V'(6)dr \\ &=36\pi(6.5-6)=18\pi \end{aligned}\] Notice that \(18\pi\) is only a fair approximation to \(19.5416\bar{6}\pi\).

   We check with an alternate computation of the differential. Remembering that \(V_{\tan}(r)=36\pi\,r-144\pi\), the differential is \[\begin{aligned} dV &=V_{\tan}(6.5)-V(6) \\ &=36\pi\,6.5-144\pi-72\pi=18\pi \end{aligned}\] The original computation of the differential was easier because we never needed to compute \(V_{\tan}(6.5)\).