Whenever a quantity $y$y is expressed as a function of another quantity $x$x and the function varies smoothly with variations in $x$x over a given domain, we can think of $y$y having a rate of change with respect to $x$x. This corresponds to the gradient of the graph of $y$y as a function of $x$x and it corresponds to the derivative of the function $y$y.
A frequently used example of this is the case of an object moving so that its distance $s$s from its starting point is given by some function of time $t$t. We can find a function that describes the speed of the object at time $t$t by differentiating the distance function. This is because speed is understood to be the rate of change of distance with respect to time.
More generally, we can often discover properties of a function by investigating the behaviour of its derivatives, including not only the first derivative but also the second and subsequent derivatives.
We find local maxima and minima and other stationary points, for example, by setting the first derivative to zero. This works because the procedure finds points where the gradient is zero.
The second derivative can help in distinguishing between types of stationary points since it expresses the rate of change of the first derivative.
If the first derivative is increasing at a stationary point, the point is a local minimum. If the first derivative is decreasing at a stationary point, the point is a local maximum. If the second derivative is zero at a stationary point, the point could be a maximum or a minimum or a point of inflection .
Examine the behaviour of the cubic polynomial function in the real number $x$x over the open interval $(-4,4)$(−4,4), defined by $p(x)=2x^3-3x^2-36x+5$p(x)=2x3−3x2−36x+5.
Begin by finding its turning points (local maximum and local minimum). Also, find its global maximum and minimum.
The derivative is $p'(x)=6x^2-6x-36$p′(x)=6x2−6x−36. Setting this to zero, we have
$0$0 | $=$= | $6x^2-6x-36$6x2−6x−36 |
$=$= | $x^2-x-6$x2−x−6 | |
$=$= | $(x-3)(x+2)$(x−3)(x+2) |
Therefore, the function is stationary at $x=3$x=3 and at $x=-2$x=−2. These could be turning points. We check the second derivative at these points. Since $p''(x)=12x-6$p′′(x)=12x−6, we have $p''(-2)=-30$p′′(−2)=−30 and $p''(3)=30$p′′(3)=30. This indicates that there is a local maximum at $x=-2$x=−2 and a local minimum at $x=3$x=3.
The values of $p$p at these points are,
$p(-2)=2\times(-2)^3-3\times(-2)^2-36\times(-2)+5=49$p(−2)=2×(−2)3−3×(−2)2−36×(−2)+5=49 and
$p(3)=2\times3^3-3\times3^2-36\times3+5=-76$p(3)=2×33−3×32−36×3+5=−76.
The only other places where there could be a maximum or a minimum are at the endpoints of the domain interval. These are at
$p(-4)=2\times(-4)^3-3\times(-4)^2-36\times(-4)+5=-27$p(−4)=2×(−4)3−3×(−4)2−36×(−4)+5=−27 and
$p(4)=2\times4^3-3\times4^2-36\times4+5=-59$p(4)=2×43−3×42−36×4+5=−59.
Thus, the global maximum is the point $(-2,49)$(−2,49) and the global minimum is the point $(3,-76)$(3,−76).
We might also note that the second derivative $p''(x)=12x-6$p′′(x)=12x−6 is zero at $x=\frac{1}{2}$x=12. This is the point at which the gradient of $p$p stops decreasing and begins to increase. It is an inflection point.
The points discussed are shown in the graph below.
The diagram illustrates the following scenario.
A lifesaver on the sand a distance $A$A from the edge of the water must reach a swimmer in the water a distance $B$B from the edge. The lifesaver's speed over sand is $V_1$V1 and over water is $V_2$V2. The relevant distances and angles are as indicated in the diagram. Clearly, the lifesaver's choice of angle $\theta_1$θ1 will determine the total time $T$T taken to reach the swimmer. Equivalently, the lifesaver can vary the distance $x$x to optimise the total time.
By Pythagoras, we have $(V_1t_1)^2=A^2+x^2$(V1t1)2=A2+x2 and $(V_2t_2)^2=B^2+y^2$(V2t2)2=B2+y2. So, the total time taken is
$T=t_1+t_2=\frac{\sqrt{A^2+x^2}}{V_1}+\frac{\sqrt{B^2+y^2}}{V_2}$T=t1+t2=√A2+x2V1+√B2+y2V2
The distances in the diagram marked with a capital letter are constants and the other quantities are variable. We can also put $x+y=D$x+y=D another constant, which will allow us to write $y=D-x$y=D−x. and $\frac{\mathrm{d}y}{\mathrm{d}x}=-1$dydx=−1.
We have expressed $T$T as a function of $x$x and the standard procedure in calculus is to differentiate $T$T with respect to $x$x to look for a minimum turning point in the function.
By differentiation, we have
$\frac{\mathrm{d}T}{\mathrm{d}x}=\frac{2x}{2V_1}\left(A^2+x^2\right)^{-\frac{1}{2}}+\frac{2y}{2V_2}\left(B^2+y^2\right)^{-\frac{1}{2}}.\frac{\mathrm{d}y}{\mathrm{d}x}$dTdx=2x2V1(A2+x2)−12+2y2V2(B2+y2)−12.dydx
(The term $\frac{\mathrm{d}y}{\mathrm{d}x}$dydx comes from using the chain rule.) After some simplification, we have
$\frac{\mathrm{d}T}{\mathrm{d}x}$dTdx | $=$= | $\frac{x}{V_1\sqrt{A^2+x^2}}-\frac{y}{V_2\sqrt{B^2+y^2}}$xV1√A2+x2−yV2√B2+y2 |
$=$= | $\frac{\sin\theta_1}{V_1}-\frac{\sin\theta_2}{V_2}$sinθ1V1−sinθ2V2 |
Following the usual practice of setting the derivative to zero, we find
$V_2\sin\theta_1=V_1\sin\theta_2$V2sinθ1=V1sinθ2 or
$\frac{V_1}{V_2}=\frac{\sin\theta_1}{\sin\theta_2}$V1V2=sinθ1sinθ2
This is the condition that minimises the total time $T$T.
The same relation is found in physics concerning the refraction of light passing from one medium to another, where it is known as Snell's Law.
The position (in metres) of an object along a straight line after $t$t seconds is modelled by $x\left(t\right)=3t^2+5t+2$x(t)=3t2+5t+2.
We want to find the velocity of the object after $4$4 seconds.
Determine $v\left(t\right)$v(t), the velocity function.
What is the velocity of the object after $4$4 seconds?
Consider the function $f\left(x\right)=2x^3-12x^2+18x+3$f(x)=2x3−12x2+18x+3.
Solve for the $x$x-coordinates of the turning points of the function.
Determine $f''\left(1\right)$f′′(1).
Determine $f''\left(3\right)$f′′(3).
State the coordinates of the local maximum.
State the coordinates of the local minimum.
State the absolute maximum value of the function within the range $0\le x\le7$0≤x≤7.
State the absolute minimum value of the function within the range $0\le x\le7$0≤x≤7.
In the diagram below, the function $y=\frac{b}{x^2+a}$y=bx2+a has a maximum turning point at $\left(0,4\right)$(0,4) and passes through $\left(3,1\right)$(3,1).
A rectangle $BCDE$BCDE is inscribed within the curve as shown. $OY$OY is the axis of symmetry.
Using the fact that the function passes through $\left(0,4\right)$(0,4) find an expression for $b$b in terms of $a$a.
Using the results of part (a) and the fact that the function passes through $\left(3,1\right)$(3,1), find the value of $a$a.
Using the results of part (a) and (b), find the value of $b$b.
If $C$C has coordinates $\left(p,0\right)$(p,0), find the $y$y coordinate of $B$B in terms of $p$p.
Find the area $A(p)$A(p) of $BCDE$BCDE in terms of $p$p.
Find the $p$p values for all turning points of $A\left(p\right)$A(p). Write each solution on the same line, separated by commas.
You may use the substitutions $u=24p$u=24p and $v=p^2+3$v=p2+3.
By filling in the blanks, show that $p=\sqrt{3}$p=√3 results in the maximum area of $BCDE$BCDE, and find that maximum area. Round any necessary answers to two decimal places.
$p$p | $1$1 | $\sqrt{3}$√3 | $2$2 |
---|---|---|---|
$A'\left(p\right)$A′(p) | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Maximum area: $A\left(\sqrt{3}\right)=\editable{}$A(√3)= units^{2}
Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods
Apply differentiation methods in solving problems