# Differentials

Consider a function $y=ƒ(x)$. Then

1. the differential, $dx$, of the independent variable $x$, is an arbitrary increment of $x$ i.e. $dx=\Delta x$
2. the differential, $dy$ of the dependent variable $y$ is $dy=ƒ'(x)dx$ The differential $dy$ of the dependent variable is not, in general, equal to the corresponding increment $\Delta y$. Because $\Delta y=ƒ(x+\Delta x)-ƒ(x)$ $\text{and, } dy=ƒ'(x)dx$

Let $P(x,y)$ and $Q(x+\Delta x,y+\Delta y)$ be two neighbouring points on the graph of $y=ƒ(x)$.

We have, $\text{Slope at point P}=ƒ'(x)$ $\tan\theta=ƒ'(x)$ $\frac{TR}{PR}=ƒ'(x)$ $\therefore dy=ƒ'(x)dx$ $TR=dy$ $QR=\Delta y$ Difference between $\Delta y$ and $dy$ is $TQ$. $TQ$ can be made very small by taking $\Delta x$ sufficiently small, so that $dy$ will approximate $\Delta y$.

This is also known as the tangent line approximation as $dy$ is the tangent line increment and $\Delta y$ is the curve increment for an increment $\Delta x$ or $dx$ in $x$.

### Compute $\Delta y$, $dy$ and $\Delta y-dy$ when $y=\frac{x^2}{2}+3x$, $x=2$ and $dx=0.5$.

Here, $y=\frac{x^2}{2}+3x$ $\therefore dy=\left(\frac{2x}{2}+3\right)dx$ $dy=(x+3)dx$ When $x=2$ and $dx=0.5$, $dy=(2+3)×0.5$ $dy=2.5$ Again, $\Delta y=ƒ(x+\Delta x)-ƒ(x)$ $=\frac{(x+\Delta x)^2}{2}+3(x+\Delta x)-\frac{x^2}{2}-3x$ $=\frac{x^2+2x.\Delta x+\Delta x^2}{2}+3(x+\Delta x)-\frac{x^2}{2}-3x$ $=\frac{x^2}{2}+\frac{2x.\Delta x}{2}+\frac{\Delta x^2}{2}+3x+3\Delta x-\frac{x^2}{2}-3x$ $=x.\Delta x+\frac{\Delta x^2}{2}+3\Delta x$ When $x=2$ and $dx=0.5$, $\Delta y=2×0.5+\frac{0.5^2}{2}+3×0.5$ $=2.625$ and, $\Delta y-dy=2.625-2.5=0.125$

### Find an approximate change in the volume of a cube of side $x\text{ m}$, caused by increasing the sides by $1\%$. What is the percentage increment in the volume?

Here, side of the cube = $x\text{ m}$ and, volume of the cube $(V)=x^3\text{ m}^3$.

Approximate change in $x$, $dx=\Delta x=1\%\text{ of }x$ $dx=0.01x$ $\therefore \Delta V=(x+\Delta x)^3-x^3$ $V=(x+0.01)^3-x^3$ $V=(1.01x)^3-x^3$ $V=1.03x^3-x^3$ $V=0.03x^3\text{ m}^3$ Thus, the percentage increment in volume $=\frac{0.03x^3}{x^3}×100\%$ $=3\%$

### Use differentials to approximate the change in $x^3$ as $x$ changes from $5$ to $5.01$.

Let $y=x^3$ and $x$ changes from $5$ to $5.01$. $\therefore x=5 \text{ and } x+\Delta x=5.01$ $\therefore \Delta x=dx=5.01-5=0.01$ Approximate change in $y$, $dy=3x^2dx$ $dy=3×5^2×0.01$ $dy=0.75$

### Find an approximate change in $\frac{1}{x}$ as $x$ changes from $1$ to $0.98$.

Let $y=\frac{1}{x}$ and $x$ changes from $1$ to $0.98$. $\therefore x=1 \text{ and } x+\Delta x=0.98$ $\therefore \Delta x=dx=0.98-1=-0.02$ Approximate change in $y$, $dy=-\frac{1}{x^2}dx$ $dy=-\frac{1}{1^2}×(-0.02)$ $dy=0.02$

### A circular copper plate is heated so that its radius increases from $5\text{ cm}$ to $5.06\text{ cm}$. Find the approximate increase in area and also the actual increase in area.

Let $x$ be the radius of the circular plate and $A$ be the area. Then $A=πx^2$ $x$ increases from $5\text{ cm}$ to $5.06\text{ cm}$. $\therefore x=5 \text{ and } x+\Delta x=5.06$ $\therefore \Delta x=dx=5.06-5=0.06$ Approximate increase in area, $dA=2πxdx$ $dA=2π×5×0.06$ $dA=0.6π\text{ cm}^2$ Actual increase in area, $\Delta A=ƒ(x+\Delta x)-ƒ(x)$ $=π(x+\Delta x)^2-πx^2$ $=π[(x+\Delta x)^2-x^2]$ $=π[5.06^2-5^2]$ $=0.6036π\text{ cm}^2$

### Find the approximate increase in the surface area of a cube if the edge increases from $10$ to $10.01\text{ cm}$. Also calculate the percentage error in the use of differential approximation.

Let $x$ be the edge of the cube and $A$ be its surface area. Then $A=6x^2$ $x$ increases from $10$ to $10.01$. $\therefore x=10 \text{ and } x+\Delta x=10.01$ $\therefore \Delta x=dx=10.01-10=0.01$ Approximate change in area, $dA=12xdx$ $=12×10×0.01$ $=1.2\text{ cm}^2$ Actual change in area, $\Delta A=ƒ(x+\Delta x)-ƒ(x)$ $=6(x+\Delta x)^2-6x^2$ $=6[(x+\Delta x)^2-x^2]$ $=6[10.01^2-10^2]$ $=1.2006\text{ cm}^2$ $\therefore \text{Error}=\Delta A-dA$ $=1.2006-1.2$ $=0.0006\text{ cm}^2$ When $x=10\text{ cm}$, then, area is $6×10^2=600\text{ cm}^2$. Thus, $\text{Percentage error}=\frac{0.0006}{600}×100\%$ $=0.0001\%$

### Find the approximate increase in the volume of a sphere when its radius increases from $2$ to $2.1$. Find also the actual increase and compare the two values.

Let $x$ be the radius and $V$ be the volume of sphere. Then $V=\frac{4}{3}πx^3$ $x$ increases from $2$ to $2.1$. $\therefore x=2 \text{ and } x+\Delta x=2.1$ $\therefore \Delta x=dx=2.1-2=0.1$ Approximate increase in volume, $dV=4πx^2dx$ $=4π×2^2×0.1$ $=1.6π$ Actual increase in volume, $\Delta V=ƒ(x+\Delta x)-ƒ(x)$ $=\frac{4}{3}(x+\Delta x)^3-\frac{4}{3}πx^3$ $=\frac{4}{3}π[(x+\Delta x)^3-x^3]$$=\frac{4}{3}π[2.1^3-2^3]$ $=\frac{5.044}{3}π$ Comparison between $dV$ and $\Delta V$, $\frac{dV}{\Delta V}=\frac{1.6π}{\frac{5.044}{3}π}=0.9516$