 ## D11 Small changes and approximations Sometimes a small change in one variable can render a big change in a larger value. For example, a small increase (or error) in the radius of a sphere means a lot more volume is added! If you estimate the small error in one variable, you can calculate the significant change in a larger variable by using derivatives.

The curve of any differentiable function will appear linear if observed over a sufficiently limited interval. If $$a$$ is a point within that interval the tangent line through $$(a,f(a))$$ provides a reasonable approximation of values of $$f(x)$$ providing $$x$$ remains near $$a.$$ A function that is differentiable at a specific point $$(a,f(a))$$ has a tangent line defined by the equation

$y=f(a)+f'(a)(x-a)$

Using the tangent line to approximate a function is called

$L(x)=f(a)+f'(a)(x-a)$

#### Example:

Consider the graph of the function $$f(x)=e^{2-x}$$ near $$x=2.$$

Note that $$f'(x)=-e^{2-x}$$ The tangent line at $$x=2$$ is given by: \begin{align*} y & =f(2)+f'(2)(x-2)\\ y & =1+(-1)(x-2)\\ y & =3-x \end{align*}

For values of $$x$$ near 2 the graph of the tangent line is close to the graph of $$f(x)=e^{2-x}$$ and the tangent line can be used to approximate $$f(x).$$

If $$x=1.9$$ , $$f(1.9)=e^{(2-1.9)}\approx1.105$$. At $$x=1.9$$ on the tangent line the value of $$y$$ is $$y=3-1.9=1.1$$

If $$x=2.05$$ , $$f(2.05)\approx0.951$$. At $$x=2.05$$ on the tangent line the value of $$y$$ is $$0.95$$

#### Example:

\begin{align*} f(x)=\sqrt{x} & \Rightarrow f'(x)=\frac{x^{-\frac{2}{3}}}{3}\\ f(8)=\sqrt{8}=2 & \ and\ f'(8)=\frac{8^{-\frac{2}{3}}}{3}=\frac{1}{12}\\ L(x) & =f(a)+f'(a)(x-a)\\ L(7.98) & =2+\frac{1}{12}(7.98-8)\\ & =1.998 \end{align*}

#### Differentials

$$\triangle x$$ is used to indicate a change in $$x$$. $$\triangle y$$ is used to indicate a change in $$y$$. $$\triangle x$$ and $$\triangle y$$ are called differentials. Differentials can be used to estimate the amount a function will change as a result of a small change in $$x$$. Consider the graph of the function $$y=f(x)$$ above and the tangent to the function at the point P. As illustrated in the diagram for values of $$x$$ close to $$1$$ the slope of the tangent at $$1$$, $$f'(1),$$ is reasonably approximated by $$\frac{\triangle y}{\triangle x}$$

\begin{align*} \textrm{More generally, as}\triangle x\rightarrow0,\frac{\triangle y}{\triangle x} & \rightarrow\frac{dy}{dx}\\ \textrm{For small values of \triangle x: }\triangle y & \approx\frac{dy}{dx}\triangle x \end{align*}

#### Examples

1.  The side of a square has length $$5\,cm.$$ How much will the area of the square increase when the side length is increased by $$0.01\,cm$$?

Let the area of the square be$$A$$ and let the length of a side be$$x\,cm$$. Then $$A=x^{2}$$ and $$\frac{dA}{dx}=2x.$$ \begin{align*} \triangle A & \approx\frac{dA}{dx}\triangle x\\ & =2x\triangle x\\ & =2\times5\times0.01\\ & =0.1 \end{align*} The increase in area is approximately $$0.1\,cm^{2}$$.

1.  A $$2$$% error is made in measuring the radius of a sphere. Find the percentage error in the volume?

Let the radius be $$r$$ and the volume be $$V$$ then $$\Delta r=0.02r.$$ Since the volume of the sphere is given by $$V=\frac{4}{3}\pi r^{3},$$ $$\frac{dV}{dr}=4\pi r^{2}.$$

First find $$\triangle V:$$ \begin{align*} \triangle V & \approx\frac{dV}{dr}\triangle r\\ & =4\pi r^{2}\triangle r\\ & =4\pi r^{2}\times0.02r\\ & =0.08\pi r^{3} \end{align*} The percentage error in the volume: \begin{align*} \textrm{Percentage error in V } & \approx\frac{\Delta V}{V}\times100\%\\ & =\frac{0.08\pi r^{3}}{\frac{4}{3}\pi r^{3}}\times100\%\\ & =\frac{0.08}{\frac{4}{3}}\times100\%\\ & =6\%. \end{align*} The percentage error in the volume is approximately 6%.

#### Exercises

1. Use linear approximation of the function $$f(x)=\sqrt{x}$$ to find the approximate value of $$\sqrt{16.1}$$

2. Use linear approximation to find an approximate value of $$\cos59^{\circ}.$$ (Hint: convert degrees to radians: $$1^{\circ}=\frac{\pi}{180}\,Radians.$$)

3. If the radius of a sphere is increased from $$10\,cm$$ to $$10.1\,cm$$ what is the approximate increase in surface area?

4. The height of a cylinder is $$10\,cm$$ and its radius is $$4\,cm$$. Find the approximate increase in volume when the radius increases to $$\text{4.02}\,cm$$.

5. An error of 3% is made in measuring the radius of a sphere. Find the percentage error in volume.

Hint: useful formula

Surface area of sphere , $$A=4\pi r^{2}$$

Volume of cylinder , $$V=\pi r^{2}h$$

Volume of sphere , $$V=\frac{4}{3}\pi r^{3}$$

1.    4.0125

2.     0.515

3.     25.13 cm$$\textrm{^{2} }$$

4.     5.03cm$$^{3}$$

5.     9%

What's next... D12 Implicit differentiation

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