So, assuming you arrived here from Section 1, you may well be asking “how do we find the gradient of the graph when the lines are not nice and straight?”
For example, when Galileo investigated motion by rolling balls down inclined planes (slopes), he wrote in his famous ‘Discorsi’ that “the spaces traversed were to each other as the squares of the times”. That means that the distance travelled by the falling object is proportional to the ‘elapsed-time-squared’.
Galileo’s experiments on motion are probably most famous for disproving Aristotle’s assertion that the acceleration of falling bodies (‘bodies’ just means ‘objects’) is proportional to their mass. Aristotle’s idea was so plausible (a hammer does indeed fall faster than a feather on Earth) that it remained scientific currency for nearly two millennia. It isn’t true, though. Hammers only fall faster than feathers because the air resistance on a falling feather is significant enough, in comparison to its weight, to significantly retard its fall. The same isn’t true of hammers, for which the air resistance is too small to have a large effect on the motion, at least at the speeds involved in dropping it from a sensible height. Indeed, a small hammer and a large hammer don’t fall appreciably differently. If you can get rid of the air resistance, then the feather would fall like the hammer, as the Apollo 15 astronauts demonstrated:
But, for our purposes, it is the proportionality of distance travelled to the square of the time elapsed that interests use here. In other words, we’ll look at hammers in this Section, not feathers…
Mathematically, we write the proportionality as:
[latex]\displaystyle x\propto t^2[/latex]
The ‘[latex]\propto [/latex]’ symbol means ‘proportional to…’. If two quantities have a proportional relationship then, logically enough, they ‘vary in proportion’ to each other. That means that if [latex]A[/latex] is proportional to [latex]B[/latex], and [latex]B[/latex] doubles, then [latex]A[/latex] will double too. If [latex]B[/latex] decreases by a factor of 1.783, then so will [latex]A[/latex]. In our example, if [latex]t[/latex] doubles, then [latex]x[/latex] will increase by a factor of 4 (because of the squared sign). The physical reason for this is that the object is accelerating, so it goes further in the second half of the motion than the first (whereas for non-accelerated motion – then [latex]x \propto t[/latex], so travelling for twice as long results in twice the distance travelled, as you may well expect). Constant acceleration, like the simplified freefall under gravity that we are considering, gives the result above.
Likewise, if two quantities are ‘inversely proportional’ means that as one, say, increases by a factor of 1.783, the other gets smaller by that factor. If [latex]A[/latex] is inversely proportional to [latex]B[/latex], then this can be written [latex]A \propto \frac{1}{B}[/latex]
The gradient of the curve [latex]x=4.9t^2[/latex]
First, we should say that the ‘4.9’ in this subheading is what is called a ‘constant of proportionality’. The behaviour of the relationship between [latex]x[/latex] and [latex]t[/latex] is contained in the [latex]x \propto t^2[/latex] part and you could sketch a graph of this the right shape (without numbers on the axes), but to make calculations (or to put numbers on the graph axes) it needs to be ‘scaled’ correctly, so that putting in a value of [latex]t[/latex] ‘spits out’ the correct value of [latex]x[/latex]. This ‘scaling’ is the job of the ‘4.9’ in the equation.
This relationship is only true (a) near the Earth’s surface and (b) if air resistance is negligible (so small that you can ignore it), but hey, how else are you going to drop a hammer?
In the paragraph above, we ‘inputted’ a number ([latex]t[/latex]), something happened to it (it got squared and then multiplied by ‘4.9’), and we were presented with an ‘output’ ([latex]x[/latex]). This is an example of a mathematical function. We can say that:
- ‘[latex]x[/latex] is a function of [latex]t[/latex]’. We write this as [latex]x=f(t)[/latex]. When you see [latex]A(B)[/latex] written, it denotes ‘[latex]A[/latex], which is a function of [latex]B[/latex]…’. Likewise [latex]A(B,C)[/latex] means ‘[latex]A[/latex], which is a function of [latex]B[/latex] and [latex]C[/latex]…
- The function for [latex]x[/latex] is [latex]x=4.9t^2[/latex]
We’ll see more about functions of more than one variable, like [latex]A(B,C)[/latex], in Section 2b
A displacement-time graph ([latex]x[/latex] against [latex]t[/latex]) of this motion would look as shown in the screenshot of the following video. How do we analyse the velocity of this motion? After all, the line isn’t straight like it was in the previous section, so how do we find the gradient of the line, in order to get the ‘rate of change’? Well, watch the video to find out…
You could try this ‘first principles’ method yourself to find the rate of change of [latex]x[/latex] as a function of [latex]t[/latex] when [latex]x=t^3[/latex]. Don’t worry – you won’t have to do this ever again – there are shortcuts…
Differentiation
Now, look at what just happened. We started by finding a ‘rate of change’ over a time interval, and then found an ‘instantaneous rate of change’ by drawing a tangent to the curve. Finally, we replaced the geometrical approach of drawing a tangent to the curve and finding its gradient with an algebraic approach, which we call ‘differentiation’.
The result of a differentiation is called a ‘derivative’, so we can say that velocity, [latex]v[/latex], is the ‘derivative with respect to time’ of displacement. The mathematical notation for a derivative with respect to [latex]t[/latex] is [latex]\frac{d}{dt}[/latex], so that [latex]v=\frac{dx}{dt}[/latex], which is usually pronounced “dee-ex-by-dee-tee”.
We also know that acceleration is the rate of change of velocity, so that [latex]\textstyle a=\frac{dv}{dt}[/latex].
Now, [latex]v[/latex] itself is the rate of change of displacement, which makes acceleration the ‘rate of change of the rate of change of displacement’, or the ‘second derivative’ of displacement.
Mathematically. We can write this as [latex]\textstyle a=\frac{d}{dt}(\frac{dx}{dt})[/latex]
Obviously, it can get cumbersome to write this all the time, so there is a shorthand for the second derivative, [latex]\textstyle \frac{d^2x}{dt^2}[/latex], which is pronounced “dee-two-ex-by-dee-tee-squared”
Third derivatives work the same:
[latex]\displaystyle \frac{d}{dt}(\frac{d^2x}{dt^2})=\frac{d^3x}{dt^3}[/latex]
The ‘nth’ derivative of [latex]x[/latex] with respect to [latex]t[/latex] is shown as [latex]\frac{d^nx}{dt^n}[/latex]
People like lift (elevator) manufacturers care about the quantity [latex]\textstyle \frac{d^3x}{dt^3}[/latex] because it relates to the ‘smoothness’ of motion as sensed by a person in the lift (elevator). It’s sometimes called ‘jerk’.
A note on ‘limits’
We said in the video that we were finding the gradient of the tangent in the limit of shrinking the tangent line to 0 size. Why do we bother with this ‘in the limit’ stuff? Why not say ‘when the tangent line has zero size? The answer is because the gradient would then be 0÷0, which is not defined in mathematics. Instead, we imagine shrinking the tangent line from a finite size to as small as our imagination can deal with (but not quite zero!) and we see that ‘in the limit as the time interval approaches zero’, the gradient successively comes closer and closer to a certain value… which is the value we want.
In fact we can write the derivative of [latex]x[/latex] with respect to [latex]t[/latex] as:
[latex]\displaystyle \frac{dx}{dt}=\lim_{x \rightarrow 0}\frac{x(t+\Delta t)-x(t)}{\Delta t}[/latex]
Can you see how this mathematical definition fits to the preceding paragraph? Take a minute or two if you need to…
So why differentiate?
What are the advantages of differentiating, rather than drawing tangents? Well, three good ones to start with are:
- The tangent approach will always be approximate (to be honest, you’ll never get the same answer twice…), whereas differentiation is exact
- If you want a velocity at a different time, you have to draw a whole new tangent, which is inconvenient to say the least. Also, you have to draw the graph in the first place… On the other hand, differentiation gives you an algebraic expression for the velocity as a function of time. You then enter the time into this function, and it ‘spits out’ the velocity.
- You can get ‘rates of change of rates of change’ by differentiating twice, whereas you’ll find that quite tricky graphically!
Having said that, you might think that the algebra we had to go through in the video was only marginally more convenient than drawing graphs and tangents. But we only showed you that algebra so that you could see where the result comes from – you don’t need to do it every time. Now we know that the derivative of [latex]t^2[/latex](with respect to [latex]t[/latex]) is [latex]2t[/latex], then every time we want to differentiate [latex]t^2[/latex], we can just write [latex]2t[/latex] as the answer. Admittedly, most of the time we differentiate, we will want to differentiate other functions, but they each have their own derivative, and there are common ones that you can memorise (we give some examples at the bottom of the page).
There is another notation for expressing derivatives with respect to time, and that involves putting a dot on top of the variable. So [latex]\dot{x}[/latex] is shorthand for [latex]\textstyle \frac{dx}{dt}[/latex], and [latex]\ddot{x}[/latex] means [latex]\textstyle \frac{d^2x}{dt^2}[/latex]. We will use this alternative notation a lot later on, especially in Sections 15 – 18, when we look at oscillations and waves.
So far we have just looked at differentiating with respect to time, but just like in Section one where we said that ‘rates of change’ could be with respect to things other than time, we can have derivatives with respect to other variables. So for example, if temperature [latex]T[/latex] varies over a distance [latex]x[/latex], then the instantaneous rate of change of temperature with distance (the temperature gradient) is [latex]\frac{dT}{dx}[/latex], measured typically in ºC per metre (ºC m-1), whereas an average temperature gradient over a distance [latex]\Delta x[/latex] would be [latex]\frac{\Delta T}{\Delta x}[/latex] (it is important you understand the difference here, so take some time to think about this if you need to)
Some examples of common derivatives are given below. Note that in Maths books, differentiation is usually taught with respect to the variable [latex]x[/latex]. This is simply because [latex]x[/latex] is the most commonly used symbol for an unknown or a variable. So in the examples that follow, [latex]x[/latex] does not mean ‘displacement’ necessarily – it is just the symbol for whatever you do mean:
[latex]\frac{d}{dx}(ax^n)=anx^{n-1}[/latex] (in other words, any power of [latex]x[/latex] can be differentiated by multiplying by the power and then reducing the power by 1). A very common example is that [latex]\frac{d}{dx}(ax)=a[/latex], so that differentiating a constant times a variable just leaves you with the constant. We will see this again and again.
[latex]\frac{d}{dx}(\sin x)=\cos x[/latex]
[latex]\frac{d}{dx}(e^x)=e^x[/latex]
[latex]\frac{d}{dx}(some\;complicated\;function\;of\; x)=\;some\;other\;complicated\;function\;of\;x[/latex]
We’ll discuss the result [latex]\frac{d}{dx}(e^x)=e^x[/latex] more in Section 5 of the site. This is a (the only) function whose derivative is equal to itself, which turns out to have particular physical significance.
A point of notation: if [latex]y[/latex] is a function, [latex]f[/latex],of [latex]x[/latex] – that is, [latex]y=f(x)[/latex], then the derivative [latex]\frac{dy}{dx}[/latex] can be (and often is) written [latex]f'(x)[/latex].
And finally, apologies for use of the word ‘Maths’ rather than ‘Math’. We are from the UK – we still write ‘colour’ and ‘honour’, after all, so go easy on us – we just can’t help it…
There are lots of techniques for differentiating various complicated functions. But when you are having to use them, don’t get bogged down in the technique, and remember that all you are doing is finding a rate of change. Don’t let the fact it might be mathematically more tricky put you off from remembering what you are trying to do in the first place.
Finally, you may be wondering, if all this was just for hammers, how do feathers fall? Well, we’ll give a brief introduction to this below, before pointing you towards a more comprehensive answer…
What effect does air resistance have on the motion of falling bodies?
We’ll use the interactive thing below to introduce the effect of air resistance, which will be a recurring theme over the next few Sections. The dashed line shows the familiar motion for freefall with no air resistance, and the solid line shows the motions when there is ‘some’ air resistance. You can change the amount of air resistance using the slider.
Really, it’s the ratio of air resistance to weight that you’re changing, but you get the picture…
INSERT INTERACTIVE ELEMENT
Notice that the gradient of the line ‘with’ air resistance is always less than the gradient ‘without’ (because it’s velocity is lower). Also, the gradient of the dashed line increases ‘forever’, whereas if you add enough air resistance you should be able to get the gradient of the solid line to remain constant after a certain point. This constant gradient means:
(a) equal distances travelled in successive equal time intervals: [latex]\frac{\Delta x}{\Delta t}=constant[/latex],
(b) constant velocity: [latex]v=constant[/latex], or [latex]\frac{dv}{dt}=0[/latex] (if something is constant, then it isn’t changing, so its rate of change is zero)
(c) zero acceleration, [latex]a=\frac{dx}{dt}=\frac{d^2x}{dt^2}=0[/latex],
which are three ways of saying the same thing. This kind of motion, in which a resistive force leads to a maximum speed (as in the case with all vehicles, where there is a ‘fight’ between the accelerating force from the engine and the resistive forces of air resistance and friction), is sometimes called ‘terminal velocity’.
Section 4, “differential equations”, looks at this type of motion in a bit more detail. And yeah, we know feathers ‘flutter’, from left-to-right-and-back-again, but ours won’t. They’ll fall straight down (just slower than hammers), and reach terminal velocity. If this lack of realism worries you, just replace the word “feather” throughout this website with a very low-density, spherical object, such as a ping-pong ball, and do your PhD on feather dynamics later…
So why didn’t we use differentiation for the car’s motion in Section 1?
Well there are two answers to this question, and you can choose your favourite:
You weren’t ready yet. The whole point of Section 1 was to lead toward Section 2…
We did use differentiation, but we just didn’t call it that.
Whichever answer you prefer, let’s revisit the car now in the following video, armed with our new knowledge…
So on this page you started ‘calculus’. The technique from calculus that you saw was ‘differentiation’, which finds instantaneous rates of change (gradients of tangents to curves).
We saw in Section 1 that the ‘inverse operation’ to finding a gradient of a section of a line, was finding the area under curve (with one important caveat that we’ll come two in Section 3).
In the same way, the inverse operation of differentiation is a thing called integration: that’s what Section 3 is about.
Link to Section 3 – Integration
If you’re new to calculus, this would be the most sensible place to go next…
Link to Home Page
And if you don’t like our suggestions, just go back here…