Approximations using Differentials

MATApproximations

Differentials let you estimate the value of a function near a known point using the linear approximation $f(x+\Delta x)\approx f(x)+f'(x)\,\Delta x$ . You pick a convenient base value $x$ where $f(x)$ is easy, take a small increment $\Delta x$ , and approximate the change $\Delta y$ by the differential $dy=f'(x)\,dx$ .

This gives quick approximate values of roots and powers, and estimates of errors in measured quantities — a short, formula-driven exam item.

Differential of y

dy=f'(x)\,dx

dx=\Delta x

is the small change in

x

;

dy

approximates the actual change

\Delta y

y

Linear approximation

f(x+\Delta x)\approx f(x)+f'(x)\,\Delta x

Choose

x

f(x)

and

f'(x)

are exact and

\Delta x

is small; valid only for small

\Delta x

Relative and percentage error

\text{rel. error}=\dfrac{dy}{y},\qquad \text{percentage error}=\dfrac{dy}{y}\times 100

dy

is the approximate error in

y

caused by the measurement error

dx

x

Cube volume error

V=x^{3}\ \Rightarrow\ dV=3x^{2}\,dx

x

is the measured edge and

dx

the possible error;

\dfrac{dV}{V}=3\dfrac{dx}{x}

gives the relative error.

Steps: pick a base $x$ with an exact value, set $\Delta x$ as the small difference, compute $f'(x)$ , then add $f(x)+f'(x)\,\Delta x$ .
For $\sqrt{25.3}$ take $x=25,\ \Delta x=0.3,\ f(x)=\sqrt{x}$ , giving $f'(x)=\dfrac{1}{2\sqrt{x}}$ .
For $(1.999)^5$ take $x=2,\ \Delta x=-0.001$ ; note $\Delta x$ is negative.
For $(255)^{1/4}$ take $x=256,\ \Delta x=-1$ since $256^{1/4}=4$ is exact.
The percentage error in $x^n$ is $n$ times the percentage error in $x$ ; so a cube's volume error percentage is $3$ times the edge's.
Differentials give an APPROXIMATE value — accuracy degrades as $|\Delta x|$ grows, so keep the increment small.
Round only at the final step to the required number of decimal places.

Where the marks go

Choosing a base point $x$ whose function value is not exact, defeating the purpose of the approximation.
Getting the sign of $\Delta x$ wrong (e.g. treating $1.999$ as $x=2$ with $\Delta x=+0.001$ instead of $-0.001$ ).
Confusing absolute error $dy$ with relative/percentage error $\dfrac{dy}{y}$ .
Using a $\Delta x$ that is too large, so the linear estimate is no longer reliable.

How the board asks it

Numericallinear approximation of roots
Using differentials, find the approximate value of $\sqrt{36.6}$ , correct to three decimal places.
Numericalapproximating a power with $x=256,\ \Delta x=-1$
Use differentials to find the approximate value of $(255)^{1/4}$ , given that $256^{1/4}=4$ .
Numericaldifferential $dy=f'(x)\,dx$ as approximate change in $y$
If $y=x^4-10$ and $x$ changes from $2$ to $1.99$ , use differentials to find the approximate change in $y$ .
Applicationapproximate error in a derived quantity
The radius of a sphere is measured as $9\ \text{cm}$ with an error of $0.03\ \text{cm}$ . Find the approximate error in calculating its surface area.
Applicationapproximate increase and percentage error
If the radius of a circle increases from $5\ \text{cm}$ to $5.1\ \text{cm}$ , use differentials to find the approximate increase in its area, and hence the approximate percentage error.

Practise this topic

PreviousRolle’s Theorem and Lagrange’s Mean Value Theorem

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.