The following exercises demonstrate how the knowledge of the derivative of logs is applied to mathematical curve sketching and to the modelling of a mathematical investigation.
Sketch the function given by $y=\frac{x^2}{\ln x}$y=x2lnx commenting on its domain and range, any limiting behaviour, discontinuities and stationary points. Prove that the tangent that has its point of contact at $x=e$x=e passes through the origin.
The domain of the curve given by $y=\frac{x^2}{\ln x}$y=x2lnx is restrained because of the presence of the log term $\ln x$lnx in the denominator. It's presence implies two things. Firstly, because the argument of a log function is non-negative, we have that $x>0$x>0. Secondly, because the term $\ln x$lnx sits in the denominator, we must have $\ln x\ne1$lnx≠1.
Thus the domain of the function becomes $x:x\in\Re^+,x\ne1$x:x∈ℜ+,x≠1.
When x is within the interval $0
As $x\rightarrow1$x→1 from the left, $\left|\ln x\right|$|lnx| becomes very small, so that $\frac{x^2}{\ln x}\rightarrow-\infty$x2lnx→−∞. With a similar argument, as $x\rightarrow1$x→1 from the right, $\frac{x^2}{\ln x}\rightarrow\infty$x2lnx→∞.
In other words, we know that there is a discontinuity at $x=1$x=1. As x approaches 1, the curve appears to approach both +\infty and -\infty at the same time! We say here that the curve approaches the vertical line $x=1$x=1 from both sides becoming asymptotic (in different directions) to that line.
As $x\rightarrow0$x→0, $\ln x\rightarrow-\infty$lnx→−∞, however this happens at a much slower rate than the numerator term $x^2\rightarrow0$x2→0.The overall effect is that the function $\frac{x^2}{\ln x}\rightarrow0$x2lnx→0.
A similar thing happens when $x$x becomes very large. The rate that the numerator $x^2$x2 increases is quicker than the rate that $\ln x$lnx increases, and so we find that the function $\frac{x^2}{\ln x}\rightarrow\infty$x2lnx→∞.
By using the quotient rule we can determine the derivative:
$y$y | $=$= | $\frac{x^2}{\ln x}$x2lnx |
$\therefore$∴ $\frac{dy}{dx}$dydx | $=$= | $\frac{\ln x\left(2x\right)-x^2\left(\frac{1}{x}\right)}{\left(\ln x\right)^2}$lnx(2x)−x2(1x)(lnx)2 |
$=$= | $\frac{2x\ln x-x}{\left(\ln x\right)^2}$2xlnx−x(lnx)2 | |
$=$= | $\frac{x\left(2\ln x-1\right)}{\left(\ln x\right)^2}$x(2lnx−1)(lnx)2 | |
Setting $\frac{dy}{dx}=0$dydx=0, we know $x\left(2\ln x-1\right)=0$x(2lnx−1)=0 and thus $x=0$x=0 or $2\ln x-1=0$2lnx−1=0.
We can discount $x=0$x=0, since the function is not defined there, so we determine that there is a stationary point where $\ln x=\frac{1}{2}$lnx=12, or in other words where $x=e^{\frac{1}{2}}=\sqrt{e}$x=e12=√e.
At $x=\sqrt{e}$x=√e, we have $y=\frac{\left(\sqrt{e}\right)^2}{\ln\sqrt{e}}=\frac{e}{\frac{1}{2}\ln e}=2e$y=(√e)2ln√e=e12lne=2e, so the stationary point has the coordinates $\left(e,2e\right)$(e,2e).
By checking either side of $x=e$x=e, we note that gradient function changes sign from negative to positive, and thus conclude that the stationary point is a local minimum turning point. This conclusion makes sense from the discussion o the functions limiting behaviour.
Thus the only stationary point is a local minimum at $\left(e,2e\right)$(e,2e)
At $x=e$x=e, $y=\frac{e^2}{\ln e}=e^2$y=e2lne=e2 and the gradient of the tangent becomes, from the derivative, $m=\frac{2e\ln e-e}{\left(\ln e\right)^2}=e$m=2elne−e(lne)2=e.
Using the point-gradient form of a line, the equation of the tangent is determined as follows:
$y-y_1$y−y1 | $=$= | $m\left(x-x_1\right)$m(x−x1) |
$y-e^2$y−e2 | $=$= | $e\left(x-e\right)$e(x−e) |
$y-e^2$y−e2 | $=$= | $ex-e^2$ex−e2 |
$\therefore$∴ $y$y | $=$= | $ex$ex |
This means that the $y$y intercept of the tangent is $0$0 and thus the tangent passes through the origin.
Here is the sketch of the function. The minimum turning point and the tangent's point of contact have been highlighted together with the asymptote $x=1$x=1.
We also observe that the range of the function is given by $y:y<0\cup y\ge2e$y:y<0∪y≥2e. That is to say it excludes values of $y$y that are both greater than or equal to $0$0 and less than $2e$2e.
Last week, Sally was investigating the harmonic sequence. This is the sequence of the reciprocals of positive integers. Mathematically we can write the sequence as $H=1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},...$H=1,12,13,14,15,...
At some stage in her investigation, she began to wonder about a new sequence, constructed by adding up what she referred to as the partial sums of $H$H as shown in the following table:
Sum | The Partial Sum | Total |
---|---|---|
$S_1$S1 | $1$1 | $1$1 |
$S_2$S2 | $1+\frac{1}{2}$1+12 | $1.5$1.5 |
$S_3$S3 | $1+\frac{1}{2}+\frac{1}{3}$1+12+13 | $1.8333...$1.8333... |
$S_4$S4 | $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$1+12+13+14 | $2.08333...$2.08333... |
$S_5$S5 | $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$1+12+13+14+15 | $2.28333...$2.28333... |
She became curious about the sequence that these sums formed (the first five terms are shown vertically in the right hand column).
It was quite obvious that the sequence was always increasing (relatively quickly at first but then the rate of increase gradually slowing down), but she wondered whether there was some mathematical function that could model it.
That is to say she was curious to find out whether or not there was a function $f\left(x\right)$f(x) that could be used to determine the approximate value of the general partial sum given by $S_n=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n}$Sn=1+12+13+14+...+1n whenever $x$x was any positive integer $n$n.
Using a spreadsheet she produced the following graph of the first $100$100 partial sums:
After a little trial and error, Sally came up with a modelling function involving natural logarithms. Her function was given by $f\left(x\right)=0.9559\ln x+0.7622$f(x)=0.9559lnx+0.7622, where whole values of $x$x corresponded to values of $n$n in $S_n$Sn.
Sally decided to construct a table listing $10$10 calculated function values $f\left(x\right)$f(x) along side the corresponding values of $S_n$Sn taken from the spreadsheet.
Specifically, she listed the results for $x=10,20,30,...,100$x=10,20,30,...,100 as shown here:
$x,n$x,n | $f\left(x\right)$f(x) | $S_n$Sn |
---|---|---|
$10$10 | $2.9632$2.9632 | $2.9290$2.9290 |
$20$20 | $3.6258$3.6258 | $3.5977$3.5977 |
$30$30 | $4.0134$4.0134 | $3.9950$3.9950 |
$40$40 | $4.2884$4.2884 | $4.2785$4.2785 |
$50$50 | $4.5017$4.5017 | $4.4992$4.4992 |
$60$60 | $4.6759$4.6759 | $4.6799$4.6799 |
$70$70 | $4.8233$4.8233 | $4.8328$4.8328 |
$80$80 | $4.9510$4.9510 | $4.9655$4.9655 |
$90$90 | $5.0636$5.0636 | $5.0826$5.0826 |
$100$100 | $5.1643$5.1643 | $5.1874$5.1874 |
As an example of Sally's calculations, to four decimal places, $f\left(20\right)=0.9559\left(\ln20\right)+0.7622=3.6258$f(20)=0.9559(ln20)+0.7622=3.6258.
The log function models the partial sums fairly closely.
Sally then determined a measure of the rate of change of the partial sums, given by the derivative $f'\left(x\right)=0.9559\left(\frac{1}{x}\right)=\frac{0.9559}{x}$f′(x)=0.9559(1x)=0.9559x .
As the number of terms in the sum becomes larger and larger, the rate of change steadily decreases. For example, with $x=10$x=10, $f'\left(10\right)=\frac{0.9559}{10}=0.09559$f′(10)=0.955910=0.09559, and for $x=100$x=100, $f'\left(100\right)=\frac{0.9559}{100}=0.009559$f′(100)=0.9559100=0.009559.
She could also use her formula to determine the approximate point at which the partial sum will exceed any number $k$k.
$0.9559\ln x+0.7622$0.9559lnx+0.7622 | $>$> | $k$k |
$0.9559\ln x$0.9559lnx | $>$> | $k-0.7622$k−0.7622 |
$\ln x$lnx | $>$> | $\frac{k-0.7622}{0.9559}$k−0.76220.9559 |
$x$x | $>$> | $e^{\frac{k-0.7622}{0.9559}}$ek−0.76220.9559 |
So for example if $k=6$k=6, then, based on Sally's formula, the number of terms needed to exceed $6$6 would be given by $x>e^{\frac{6-0.7622}{0.9559}}=239.71$x>e6−0.76220.9559=239.71. In other words, $240$240 terms.
A quick check on her spreadsheet showed that the actual number of terms required is $227$227, where $S_{227}=6.0044$S227=6.0044. Her estimate showed $S_{240}=6.0599$S240=6.0599.
We want to find the gradient of the curve $y=\ln\left(x^2+5\right)$y=ln(x2+5) at the point where $x=3$x=3.
Find the derivative of $y=\ln\left(x^2+5\right)$y=ln(x2+5). You may use the substitution $u=x^2+5$u=x2+5.
Now find the value of $y'$y′ at $x=3$x=3 to find the gradient of the tangent line.
Consider the function $f\left(x\right)=\frac{\ln x}{x}$f(x)=lnxx for $x>0$x>0.
Show that the graph of $f\left(x\right)$f(x) has a stationary point at $x=e$x=e. In your working you may let $u=\ln x$u=lnx and $v=x$v=x.
Determine $f''\left(e\right)$f′′(e). In your working you may let $u=1-\ln x$u=1−lnx and $v=x^2$v=x2.
Hence determine whether the stationary point is a maximum or minimum.
Maximum
Minimum
Maximum
Minimum
A plane takes off from an airport at sea level and its altitude $h$h in metres, $t$t minutes after taking off, is given by $h=600\ln\left(t+1\right)$h=600ln(t+1).
Exactly $t$t minutes after taking off, at what rate is the plane ascending?
Hence, what is the rate of ascent at exactly $4$4 minutes after take off?
How would you describe the ascent of the plane?
Ascending, at an increasing rate.
Ascending, but at a decreasing rate.
Ascending, at an increasing rate.
Ascending, but at a decreasing rate.
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