C.3 Graphs

Numerical, algebraic and graphical forms can describe the same physical relationship. Translation means preserving the variables, units and conditions while changing how the relationship is represented.

In a linear stress-strain region with stress on the vertical axis and strain on the horizontal axis, Young modulus $E$ is the gradient. A stress of $120\,\mathrm{MPa}$ at strain $6.0\times10^{-4}$ gives $E=120\times10^6/(6.0\times10^{-4})=2.0\times10^{11}\,\mathrm{Pa}$.

A useful graph gives each measured pair an unambiguous position and uses the plotting area efficiently. For ‘extension against force’, plot extension vertically and force horizontally.

If uncertainty bars are supplied or required, draw them to the stated uncertainty in the correct direction. A best-fit line should balance the overall scatter rather than be forced through the origin or through every point.

Before interpreting the plot, verify that scale increments are uniform, every point lies within the axes, and transformed variables such as $1/x$ or $\ln x$ are labelled as the quantities actually plotted.

A relationship is linear in the plotted variables when it can be written as $y=mx+c$, where $m$ and $c$ are constants. Identifying $y$, $x$, $m$ and $c$ predicts the graph before it is drawn.

A curved relationship may become linear after a justified transformation. The transformed quantity—not the original symbol alone—must occupy the axis used in the comparison with $y=mx+c$.

For constant acceleration, $v=u+at$ predicts a straight velocity-time graph: acceleration fixes its gradient and initial velocity fixes its vertical intercept.

For a best-fit straight line, choose two well-separated points on the line; they need not be measured data points. Use a large gradient triangle, retain the sign and obtain gradient units from vertical-axis unit divided by horizontal-axis unit.

The vertical intercept is the value of $y$ when $x=0$. On a velocity-time graph it can represent initial velocity. If the plotted axis is transformed, reverse that transformation: an intercept $log_{10}A_0=1.57$ gives $A_0=10^{1.57}$.

A line through $(1.0\,\mathrm{s},5.0\,\mathrm{m\,s^{-1}})$ and $(7.0\,\mathrm{s},17.0\,\mathrm{m\,s^{-1}})$ has gradient $12.0/6.0=2.0\,\mathrm{m\,s^{-2}}$.

When a graph is linear, its gradient gives one constant rate of change across the whole interval. The physical rate follows from the quantities and units on the axes.

If a straight velocity-time line rises from $4.0\,\mathrm{m\,s^{-1}}$ at $t=1.0\,\mathrm{s}$ to $16.0\,\mathrm{m\,s^{-1}}$ at $t=7.0\,\mathrm{s}$, $a=(16.0-4.0)/(7.0-1.0)=2.0\,\mathrm{m\,s^{-2}}$.

A horizontal line has zero rate. A negative gradient gives a negative rate relative to the chosen positive direction; for velocity this is negative acceleration, not automatically a decrease in speed.

For a curved graph, the rate changes from point to point. The gradient of a tangent at the chosen point estimates the instantaneous rate there because the tangent matches the curve's local direction.

On a displacement-time graph, suppose a tangent at $t=2.0\,\mathrm{s}$ passes through convenient tangent points $(1.0\,\mathrm{s},3.0\,\mathrm{m})$ and $(3.0\,\mathrm{s},11.0\,\mathrm{m})$. The instantaneous velocity is $(11.0-3.0)/(3.0-1.0)=4.0\,\mathrm{m\,s^{-1}}$.

A longer tangent triangle reduces the percentage effect of reading uncertainty. The two calculation points belong to the tangent and need not lie on the original curve.

An average rate describes change across a finite interval; an instantaneous rate describes the rate at one particular point. On a curved graph they are generally different.

If displacement changes from $2\,\mathrm{m}$ at $1\,\mathrm{s}$ to $14\,\mathrm{m}$ at $5\,\mathrm{s}$, average velocity is $(14-2)/(5-1)=3\,\mathrm{m\,s^{-1}}$. A tangent at $5\,\mathrm{s}$ could have a different gradient.

As the interval around a point becomes smaller, its chord gradient can approach the tangent gradient when the curve is smooth. This explains why a tangent represents the local rate without requiring explicit differentiation.

The area between a curve and the horizontal axis can represent a physical quantity when multiplying the axis units produces that quantity. Its meaning must come from the model, not from geometry alone.

For straight sections, add rectangle, triangle or trapezium areas. For a curve, estimate with narrow strips or count squares. Treat area below the horizontal axis as negative when the represented quantity is signed.

For a linear capacitor voltage-charge graph rising from zero to $12\,\mathrm{V}$ at $4.0\,\mathrm{mC}$, the triangular area is $\tfrac12(12)(4.0\times10^{-3})=2.4\times10^{-2}\,\mathrm{J}$. This application is full A Level content.

A rate equation links a quantity's present value to how quickly it changes. It can be explored with graph gradients or a spreadsheet using small finite time steps, without writing derivatives or integrals.

With $x=10$, $\lambda=0.20\,\mathrm{s^{-1}}$ and $\Delta t=1.0\,\mathrm{s}$, the initial rate is $-2.0$ units $\mathrm{s^{-1}}$ and the next modelled value is $10+(-2.0)(1.0)=8.0$. Repeating the rows produces a decaying curve.

For capacitor discharge $V=V_0e^{-t/\tau}$, taking a logarithm makes voltage linear in time. The gradient reveals the time constant $\tau$, but its formula depends on the logarithm used.

If a graph of $\ln(V/V_{\mathrm{ref}})$ against $t$ has gradient $-0.250\,\mathrm{s^{-1}}$, then $\tau=-1/(-0.250)=4.00\,\mathrm{s}$.

Changing the voltage reference or stated voltage unit shifts the intercept but not the gradient, provided one consistent convention is used for every point.

A proposed exponential or power law can be tested by transforming it into a straight-line form. The transformed graph must be linear and its gradient must agree with the proposed parameter.

Radioactive decay and capacitor discharge give negative gradients on $\ln y$ against time. For $F=kx^{-2}$, a log-log graph should be straight with gradient $-2$; a measured gradient far from $-2$ does not support the inverse-square claim.

Use several transformed data points, appropriate axes and a best-fit line. Judge agreement using scatter and measurement uncertainty rather than demanding an exact textbook gradient from imperfect data.

A sketch shows qualitative shape, intercepts, turning behaviour and asymptotes implied by an equation. For positive $k$, the function family predicts the following features before any numerical scale is chosen.

Physical domains can retain only part of the mathematical graph: pressure and volume are positive, time may begin at zero, and a squared physical quantity may not use the negative-$x$ branch.

Changing a positive constant $k$ stretches the vertical scale without changing the function family. A negative $k$ reflects the graph across the horizontal axis.