3 Week 3

We start this week by looking at derivatives of derivatives and then move on to look at derivatives from the point of view of rates of change of the function with respect to the variable. This leads us to look at velocity and acceleration.

3.1 Successive Differentiation

The result of differentiating a function (i.e. taking the derivative of a function) is again a function. So we can differentiate again.

Suppose \(y = x^3 + 2x^2\). Then \[ \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^2 + 4x. \]

We may then differentiate \(\mathrm{d}y/\mathrm{d} x\) as well. This is written as \[ \frac{\mathrm{d}}{\mathrm{d} x}\frac{\mathrm{d} y}{\mathrm{d} x}\text{ or more usually }\frac{\mathrm{d}^{2}y}{\mathrm{d} x^2}. \] So in this case \(\mathrm{d}^{2}y/\mathrm{d} x^2 = 6x+4\).

This is called the second derivative of \(y\).

We can repeat, and differentiate \(\mathrm{d}^{2}y/\mathrm{d} x^2\). This gives \[ \frac{\mathrm{d}^{3}y}{\mathrm{d} x^{3}} = 6 \] and is called the third derivative of \(y\). Taking the derivative of \(\mathrm{d}^{3}y/\mathrm{d} x^{3}\) yields \[ \frac{\mathrm{d}^{4}y}{\mathrm{d} x^{4}} = 0, \] the fourth derivative of \(y\).

We can continue, but as the derivative of \(0\) is \(0\), all further derivatives of \(y\) will just be \(0\).

Example 3.1 Find \(\mathrm{d}y/\mathrm{d} x, \mathrm{d}^{2}y/\mathrm{d} x^2, \mathrm{d}^{3}y/\mathrm{d} x^{3}\) for \(y =e^x\sin(x)\).

Solution. We use the product rule. If \[ y=e^x\sin(x)\text{ then }\frac{\mathrm{d} y}{\mathrm{d} x} = e^x\cos(x) + e^x\sin(x). \] Now \[ \frac{\mathrm{d}^{2}y}{\mathrm{d} x^2} = e^x(-\sin(x)) + e^x\cos(x) + e^x\cos(x)+e^x\sin(x) = 2e^x\cos(x), \] and finally \[ \frac{\mathrm{d}^{3}y}{\mathrm{d} x^{3}} = 2(-e^x\sin(x)+e^x\cos(x)) = 2e^x(\cos(x)-\sin(x)). \]

3.1.1 Rates of Change

The derivative \(\mathrm{d}y/\mathrm{d} x\) may be considered as the rate at which \(y\) changes as a result of a change in \(x\).

For example, suppose that \(s=s(t)\) gives the distance of a particle to the origin \(O\) in metres, at a time \(t\) seconds.

Suppose that \(s(t) = 3t^3 + 2t\). So at time \(t = 0\text{s}\), the particle is at \(O\); at time \(t = 1\text{s}\), the particle is \(5\text{m}\) metres away from \(O\); at time \(t = 2\text{s}\), the particle is \(28\text{m}\) away from \(O\), etc.

If we differentiate \(s\) with respect to \(t\), we get \(\dfrac{\mathrm{d} s}{\mathrm{d} t} = 9t^2 + 2\).

This tells us that the slope of the graph of \(s\) against \(t\) is not constant but varies depending on the value of \(t\). So when \(t = 0\), the slope is \(2\), when \(t = 1\), the slope is \(11\), etc.

\(\mathrm{d} s/\mathrm{d} t\) is the rate at which \(s\) changes for a change in \(t\), or the rate at which distance changes with time, so \(\mathrm{d} s/\mathrm{d} t\) is the velocity, \(v\), of the particle in \(\text{m}\text{s}^{-1}\).

If we differentiate the velocity \(v\), we get \(\mathrm{d} v/\mathrm{d} t = 18t\). Also, since the derivative of the velocity is the second derivative of the distance, we have \[ \frac{\mathrm{d} v}{\mathrm{d} t}= \frac{\mathrm{d}^{2}s}{\mathrm{d}t^2}. \] \(\mathrm{d} v/\mathrm{d} t\) is the rate of change of velocity with time or the acceleration of the particle.

Note that \(\mathrm{d}^{3}s/\mathrm{d} t^{3}\) is not zero for \(s = 3t^3 + 2t\), so we could find it. The rate of change of acceleration is called “jerk”; higher derivatives are even less common to be considered (“snap”, “crackle”, “pop”).

Example 3.2 A body moves such that its displacement \(s\), in metres, for any instant of time \(t\) seconds is given by \[ s(t) = 0.4t-0.2t^2. \]

Show that its acceleration is constant;
Determine its velocity after \(3\) seconds and its position after \(2\) seconds.

Solution.

Since \(s(t) = 0.4t-0.2t^2\) then velocity = \(\mathrm{d} s/\mathrm{d} t = 0.4-0.4t\); acceleration = \(\mathrm{d}^{2}s/dt^2 = -0.4\). So the acceleration is always \(-0.4ms^{-2}\) which is a constant value (so does not depend on the value of \(t\)).
At \(t=3, \mathrm{d} s/\mathrm{d} t = 0.4-(0.4\times 3) = -0.8\). At \(t=2, s = 0.4\times2-(0.2\times 2^2) = 0.8-0.8 = 0\). Therefore after \(3\) seconds the velocity is \(-0.8ms^{-1}\) and after 2 seconds the particle has a displacement (distance in metres from the origin \(O\)) of \(0m\).

3.2 Maxima, Minima and points of inflection

The derivative of a function at a point gives the gradient (the slope) of that curve at the point. The gradient may be positive, negative or zero. Consider Figure 3.1.

Figure 3.1: turning points

For the graph above:

The slope from \(A\) to \(B\) is positive.
The slope at \(B\) is zero.
The slope from \(B\) to \(C\) is negative.
The slope at \(C\) is zero.
The slope from \(C\) to \(D\) is positive.

As \(x\) increases from just before \(B\) to just after, the slope of the graph changes from positive to negative. A point such as \(B\) is called a maximum or local maximum. The word “local” means that this is not the “global” maximum \(y\) value of the whole curve, but is the maximum value in the region “near \(B\)”, e.g. between \(A\) and \(C\).

As \(x\) increases from just before \(C\) to just after, the slope of the graph changes from negative to positive. A point such as \(C\) is known as a minimum or local minimum.

Points such as \(B\) and \(C\) are also sometimes called critical points or stationary points or turning points. A function can have more than one maximum or minimum along its range of \(x\) values.

3.2.1 Points of inflection

At any point on a graph where a maximum or minimum occurs, the gradient or rate of change of the curve is zero i.e. \[ \frac{\mathrm{d} y}{\mathrm{d} x}=0. \] However, the converse is not necessarily true: the gradient may be zero, but the sign of the slope of the graph may not change. Such a point is called a point of inflection.

Figure 3.2: points of inflection

In the left hand graph on Figure 3.2, the slope is positive (strictly) between \(A\) and \(B\), and likewise between \(B\) and \(C\). At \(B\) it is zero.

Similarly on the right of Figure 3.2: the slope is negative between \(D\) and \(E\), between \(E\) and \(F\), and zero at \(E\).

The points \(B\) and \(E\) are both points of inflection.

We would like to be able to find out how many maxima, minima and points of inflection a curve has without having to plot the graph. We can do this by examining the derivatives of the function.

Consider the curve of the function with the equation \[ y = x^3-\frac 32x^2-18x+10. \] We know that at a stationary point, \(\mathrm{d}y/\mathrm{d} x=0\). So by finding \(\mathrm{d}y/\mathrm{d} x\) and setting it equal to zero, we can determine the \(x\) values of the stationary points as follows: \[\begin{equation} \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^2-3x-18 \tag{3.1} \end{equation}\] Finding when \(\mathrm{d}y/\mathrm{d} x = 0\) leads to the (quadratic) equation \[\begin{equation} 3x^2-3x-18=0. \tag{3.2} \end{equation}\]

We find that the quadratic equation (3.2) has roots at \(x = 3\) and \(x = -2\). So now we know that there are two turning points, one at \(x = 3\) and one at \(x = -2\). We can find the \(y\) coordinates of these points by substituting each \(x\) value into equation (3.1). This gives \(y = -30.5\) when \(x = 3\) and \(y = 32\) when \(x = -2\).

We now know the coordinates of the turning points, but we do not know whether they are maxima, minima or points of inflection. We can often establish the type of turning point by considering the second derivative \(\mathrm{d}^{2}y/\mathrm{d} x^2\).

Consider a maximum of some function (generally).

Figure 3.3: maximum turning point

In the left hand graph on Figure 3.3, of the function \(y\), we have a local maximum. The gradient of the curve, \(\mathrm{d}y/\mathrm{d} x\), at point \(A\) will be a large positive value as the curve here is steeply sloped up from left to right. We can plot the value of the gradient at \(A\) on a graph (right hand graph above, of the function \(\mathrm{d}y/\mathrm{d} x\)). At \(B\) the slope of graph \(y\) is still positive, but less steep, so the gradient at \(B\), shown on the graph of \(\mathrm{d}y/\mathrm{d} x\) is still a positive value, but is smaller than at \(A\).

At \(C\) the gradient of graph \(y\) is zero. This is the point where the curve in graph of \(\mathrm{d}y/\mathrm{d} x\) crosses the \(x\) axis. At \(D\) the gradient of graph \(y\) is now negative, but not very steep, so on graph \(\mathrm{d}y/\mathrm{d} x\) the gradient at \(D\) is a small negative value. At \(E\) the gradient of graph \(y\) is steep and negative, so on graph \(\mathrm{d}y/\mathrm{d} x\) the point is marked as a large negative value.

Now consider the gradient of graph \(\mathrm{d}y/\mathrm{d} x\). Graph \(\mathrm{d}y/\mathrm{d} x\) is a plot of \(\mathrm{d}y/\mathrm{d} x\) against \(x\). The gradient of this graph is given by \(\mathrm{d}^{2}y/\mathrm{d} x^2\). We can see from the curve in that graph that the gradient of the curve is negative at all points. That is \(\mathrm{d}^{2}y/\mathrm{d} x^2<0\) for all \(x\) values close to the local maximum. Therefore if \(\mathrm{d}^{2}y/\mathrm{d} x^2\) is a negative number for the \(x\) value of the turning point, we can be sure the turning point is a maximum.

By a similar argument and considering the graphs below (Figure 3.4, we can show that for any turning point where, \(\mathrm{d}^{2}y/\mathrm{d} x^2>0\) we have a local minimum.

Figure 3.4: minimum turning point

Previously, we established that the curve \(y = x^3-\frac32 x^2-18x+10\) had turning points at \(x=3\) and \(x=-2\). We are now in a position to find out what kind of turning points each of these is.

\[ \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^2-3x-18\text{ therefore } \frac{\mathrm{d}^{2}y}{\mathrm{d} x^2} = 6x-3. \] When \(x = 3\), \(\mathrm{d}^{2}y/\mathrm{d} x^2=15 > 0\). So we deduce that the turning point at \(x=3\) is therefore a local minimum.

When \(x = -2\), \(\mathrm{d}^{2}y/\mathrm{d} x^2 = -15 < 0\) and so we deuce that the turning point at \(x = -2\) is therefore a local maximum.

Note that the statement \[ \text{When }x = 3, \frac{\mathrm{d}^{2}y}{\mathrm{d} x^2} = 15 \] is often written as \[ \left.\frac{\mathrm{d}^{2}y}{\mathrm{d} x^2}\right\vert_{x=3} = 15. \] Sometimes you can also see the notation \[ \frac{\mathrm{d}^{2}y}{\mathrm{d} x^2}(3) = 15. \] Similarly we can write for the other statement that \[ \left.\frac{\mathrm{d}^{2}y}{\mathrm{d} x^2}\right\vert_{x=-2} = \frac{\mathrm{d}^{2}y}{\mathrm{d} x^2}(-2) = -15. \]

There is a third possibility for the value of \(\mathrm{d}^{2}y/\mathrm{d} x^2\). It may be zero.

When \(\mathrm{d}^{2}y/\mathrm{d} x^2=0\), we may have a maximum, a minimum or a point of inflection at our turning point. We will need a further test to find out which.

If we consider the points of inflection in Figure 3.2, we can see that for the left-hand graph, the slope, \(\mathrm{d}y/\mathrm{d} x\) , is positive on both sides of the point of inflection. For the right hand graph the slope is negative on both sides of the point of inflection. For a maximum the slope always changes from positive to negative as x increases and for a minimum the slope always changes from negative to positive as \(x\) increases. This suggests a way of finding out what kind of turning point we have when \(\mathrm{d}^{2}y/\mathrm{d} x^2\) turns out to be zero.

We can substitute an \(x\) value just smaller than the value at the turning point into our equation for \(\mathrm{d}y/\mathrm{d} x\) and find if \(\mathrm{d}y/\mathrm{d} x\) is positive or negative at this \(x\) value, then we can do the same for an \(x\) value just larger than the \(x\) value of the turning point and again determine the sign of the outcome.

Then the type of turning point can be established according to the following table:

slope before turning pt	slope at turning pt	slope after turning pt	type of pt
positive	zero	negative	maximum
negative	zero	positive	minimum
positive	zero	positive	inflection
negative	zero	negative	inflection

Example 3.3 Find and identify the critical or turning points of the function \(y=x^3\).

Solution. For \(y=x^3, \mathrm{d}y/\mathrm{d} x = 3x^2\). At a turning point \(\mathrm{d}y/\mathrm{d} x=0\) and so \(3x^2=0\) and so there is a turning point of some kind at \(x=0\).

\(\mathrm{d}^{2}y/\mathrm{d} x^2=6x\) and so \[ \left.\frac{\mathrm{d}^{2}y}{\mathrm{d} x^2}\right\vert_{x=0} = 0. \] Thus we cannot identify the type of turning point from the second derivative. However, we can consider the first derivative.

For \(x=-0.1\) (just less than \(0\)) \(\mathrm{d}y/\mathrm{d} x = 3(-0.1)^2 = 0.03>0\).

For \(x = 0.1\) (just greater than \(0\)) \(\mathrm{d}y/\mathrm{d} x=3(0.1)^2 = 0.03>0\).

So from examining the table above we see that the critical point at \(x=0\) must be a point of inflection.

3.2.2 Points of inflection

The points of inflection we have considered so far are a special case where the gradient is zero at some value of x. In fact any point where the curve changes from concave to convex is called a point of inflection regardless of whether the gradient becomes zero or not.

Two points of this type are shown in the graphs on Figure 3.5.

Figure 3.5: points of inflection

Here the graph changes from concave downwards to concave upwards as \(x\) increases in the first case and from concave upwards to concave downwards in the second case. In each case the marked point is a point of inflection. But the gradient is not zero at these points!

For points such as these \(\mathrm{d}y/\mathrm{d} x\) is not zero, but \(\mathrm{d}^{2}y/\mathrm{d} x^2\) always is. We can find the position of such points by finding where \(\mathrm{d}^{2}y/\mathrm{d} x^2\) is zero. We can decide which way they bend by considering \(\mathrm{d}y/\mathrm{d} x\) as for the previous point of inflection example.

Example 3.4 Find the \(x\) coordinates of all the maxima, minima and points of inflection for the curve \[ y = 2x^3-9x^2+12x. \]

Solution. For \(y = 2x^3-9x^2+12x\), \(\mathrm{d}y/\mathrm{d} x = 6x^2-18x+12\). When \(\mathrm{d}y/\mathrm{d} x=0\) then \(6x^2-18x+12=0\) and so \(x=2\) or \(x=1\).

Now \(\mathrm{d}^{2}y/\mathrm{d} x^2 = 12x-18\), so when \(x=2\), \(\mathrm{d}^{2}y/\mathrm{d} x^2>0\) and the point is a minimum.

When \(x=1\), \(\mathrm{d}^{2}y/\mathrm{d} x^2<0\) and the point is a maximum.

Also, \(\mathrm{d}^{2}y/\mathrm{d} x^2=0\) when \(x=1.5\) and so at \(x=1.5\) here is a point of inflection (but not of zero gradient).

From examining the value of \(\mathrm{d}y/\mathrm{d} x\) just before and just after \(x=1.5\) we find that the gradient is negative on both sides of the point of inflection and thus the curves slopes down from left to right.

Figure 3.6: \(y = 2x^3-9x^2+12x\)