Problems in which differentiation is applicable are often, but not always, to do with finding maxima or minima. They can concern rates of change more generally.
Except at the endpoints of the domain, at a maximum or a minimum, the gradient of a smooth function must be zero. Otherwise, there would be a point near the supposed extreme point that exceeded it. However, there can be a point of zero gradient that is neither a maximum nor a minimum.
Oscillations are often modelled using trigonometric functions. Suppose the horizontal position of an object relative to a central point is related to time $t$t by the function $x(t)=0.5\cos2t$x(t)=0.5cos2t. We wish to find an expression for the instantaneous velocity of the object at time $t$t at all times after time $t=0$t=0. When is the velocity at a maximum in magnitude?
Velocity $v(t)$v(t) is the time rate of change of position. So, we differentiate $x(t)$x(t). Thus,$v(t)=-0.5\times2\sin2t=-\sin2t$v(t)=−0.5×2sin2t=−sin2t.
To find stationary points, we set the derivative of the velocity function to zero. That is, we want to find where the acceleration function, $a(t)$a(t), is zero. Differentiating again, we find
$0=a(t)=v'(t)=-2\cos2t$0=a(t)=v′(t)=−2cos2t and this is true when $2t=\frac{\pi}{2}+n\pi$2t=π2+nπ. Thus, $t=\frac{\pi}{4}+n\frac{\pi}{2}$t=π4+nπ2, where $n$n is an integer.
Some of these solutions represent local maxima and the others, local minima. If we check the velocity function at $t=\frac{\pi}{4}$t=π4 and $t=\frac{3\pi}{4}$t=3π4 (that is, where $n=0$n=0 and $n=1$n=1), we should be able to understand where the maximum and minimum velocity values occur.
$v(\frac{\pi}{4})=-\sin\pi2=-1$v(π4)=−sinπ2=−1
$v\left(\frac{3\pi}{4}\right)=-\sin\frac{3\pi}{2}=1$v(3π4)=−sin3π2=1
In the first case, the object is moving to the left but in the second case, it is moving to the right. Therefore, we would say the times $t=\frac{\pi}{4}+n\frac{\pi}{2}$t=π4+nπ2 with $n$n even represent minima in the velocity function and when $n$n is odd, the velocity is maximised.
But, the question asked for the times when the magnitude of the velocity is maximised. So, the answer is simply $t=\frac{\pi}{4}+n\frac{\pi}{2}.$t=π4+nπ2.
Consider the following diagram in which a trapezium is inscribed in a circle. The circle has radius $1$1 unit and the line $AB$AB in the trapezium is fixed and has length $a=\sqrt{2}$a=√2. Line $DC$DC is always parallel to $AB$AB while $\theta$θ varies from $0$0 to $\pi$π. What angle $\theta$θ gives the trapezium $ABCD$ABCD its greatest area?
On an initial examination of the set-up, we see that when $\theta=0$θ=0 the trapezium becomes a triangle. The area possibly grows as $\theta$θ increases, reaches a maximum and then begins to decrease. When $\theta$θ eventually reaches a size such that point $C$C coincides with point $B$B, the area is $0$0. Beyond this, the area again begins to grow.
The area function for the trapezium will need to be defined piecewise. Otherwise, when $C$C passes beyond $B$B, the area will become negative.
With right-angled triangle trigonometry, we see that $DC$DC has length $2\sin\theta$2sinθ. So, the average width of the trapezium is $\frac{\sqrt{2}+2\sin\theta}{2}=\frac{1}{\sqrt{2}}+\sin\theta$√2+2sinθ2=1√2+sinθ. The distance between the parallel sides is $\left|\cos\theta+\frac{1}{\sqrt{2}}\right|$|cosθ+1√2|. (We used Pythagoras to get the distance from $AB$AB to the centre.)
Therefore, the area is $A=\left(\frac{1}{\sqrt{2}}+\sin\theta\right)\left|\cos\theta+\frac{1}{\sqrt{2}}\right|$A=(1√2+sinθ)|cosθ+1√2|,
Observe that when $\theta=0$θ=0, $A=\frac{1+\sqrt{2}}{2}\approx1.21$A=1+√22≈1.21. and when $\theta=\pi$θ=π, $A=\frac{\sqrt{2}-1}{2}\approx0.21$A=√2−12≈0.21. But, somewhere between these values, the area reaches zero. In fact, from the area formula it is clear that $A=0$A=0 when $\theta=\frac{3\pi}{4}$θ=3π4.
We should now give the area function, piecewise, as
$A=\left(\frac{1}{\sqrt{2}}+\sin\theta\right)\left(\cos\theta+\frac{1}{\sqrt{2}}\right)$A=(1√2+sinθ)(cosθ+1√2) , for $0\le\theta<\frac{3\pi}{4}$0≤θ<3π4, and
$A=-\left(\frac{1}{\sqrt{2}}+\sin\theta\right)\left(\cos\theta+\frac{1}{\sqrt{2}}\right)$A=−(1√2+sinθ)(cosθ+1√2), for $\frac{3\pi}{4}\le\theta\le\pi$3π4≤θ≤π.
We could differentiate $A$A with respect to $\theta$θ separately on each part of the domain and equate the derivatives to $0$0 to look for turning points. But, it is clear that the largest area must occur somewhere within the domain $0\le\theta<\frac{3\pi}{4}$0≤θ<3π4.
For this part of the domain,
$A'$A′ | $=$= | $-\sin\theta\left(\frac{1}{\sqrt{2}}+\sin\theta\right)+\left(\frac{1}{\sqrt{2}}+\cos\theta\right)\cos\theta$−sinθ(1√2+sinθ)+(1√2+cosθ)cosθ |
$=$= | $\cos^2\theta-\sin^2\theta+\frac{1}{\sqrt{2}}\left(\cos\theta-\sin\theta\right)$cos2θ−sin2θ+1√2(cosθ−sinθ) | |
$=$= | $\left(\cos\theta-\sin\theta\right)\left(\cos\theta+\sin\theta+\frac{1}{\sqrt{2}}\right)$(cosθ−sinθ)(cosθ+sinθ+1√2) |
When $A'=0$A′=0, we have either $\cos\theta-\sin\theta=0$cosθ−sinθ=0 or $\cos\theta+\sin\theta+\frac{1}{\sqrt{2}}=0$cosθ+sinθ+1√2=0. Therefore, $\cos\theta=\sin\theta$cosθ=sinθ, which occurs in the given domain when $\theta=\frac{\pi}{4}$θ=π4, or $\cos\theta+\sin\theta+\frac{1}{\sqrt{2}}=0$cosθ+sinθ+1√2=0, which provides the solution $\theta=\frac{11\pi}{12}$θ=11π12, which is in the second part of the domain.
The area when $\theta=\frac{\pi}{4}$θ=π4 is $2$2 , which is greater than the area at the upper endpoint of the domain. Also, the area at $\theta=\frac{11\pi}{12}$θ=11π12 is $\frac{1}{4}$14 - a local maximum. So, we conclude that the maximum area is $2$2, attained when $\theta=\frac{\pi}{4}$θ=π4.
Note that the minimum area occurred at the endpoint of the domain of interest. The piecewise defined function was continuous but not smooth at this point - $\theta=\frac{3\pi}{4}$θ=3π4.
The curve $y=-5x\sin x$y=−5xsinx passes through the point $P\left(\pi,0\right)$P(π,0).
Find the equation of the tangent at point $P$P.
Consider the function $y=\frac{\sin x}{1+\cos x}$y=sinx1+cosx.
Find $\frac{dy}{dx}$dydx.
How many turning points does $y$y have?
The arm of a pendulum swings between its two extreme points $A$A to the left and $B$B to the right. Its horizontal displacement $x$x cm from the centre of the swing at time $t$t seconds after it starts swinging is given by $x\left(t\right)=16\sin3\pi t$x(t)=16sin3πt.
At what position does the pendulum start swinging?
What is the furthest it gets from the central position of its swing?
Velocity is the rate at which displacement changes over time.
State the velocity function $v\left(t\right)$v(t).
Solve for the first two times at which the pendulum comes to rest.
What is the displacement of the pendulum when it first comes to rest?
What is the displacement of the pendulum when it comes to rest the second time?
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