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 $y=ƒ(x)$.

Differentials calculus

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 $\Delta V$, \[\frac{dV}{\Delta V}=\frac{1.6π}{\frac{5.044}{3}π}=0.9516\]

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