Consider the following two applications of exponential functions.
A certain model, specifically $f\left(t\right)=1-e^{-0.008t}$f(t)=1−e−0.008t, was used to analyse the length of stay of a group of patients diagnosed with schizophrenia in a local hospital. The function $f(t)$f(t) is the proportion of the group that had been discharged at the end of $t$t days of hospitalisation.
At the beginning of the study (at $t=0$t=0), the model shows $f\left(0\right)=1-e^{-0.008\times0}=1-1=0$f(0)=1−e−0.008×0=1−1=0 people had been discharged as expected.
As time passes the proportion steadily increases and we can see that as $t\rightarrow\infty$t→∞, the quantity $e^{-0.008\times t}\rightarrow0$e−0.008×t→0, and so the function $f(t)\rightarrow1$f(t)→1 as required by the nature of the model.
The domain of the function might be specified for whole days only. However, if we consider the process as a continuous discharging of patients across time ($t\in\Re$t∈ℜ), we can define the derivative as $f'\left(t\right)=0.008e^{-0.008t}$f′(t)=0.008e−0.008t.
This derivative can never be zero because the quantity $e^{-0.008t}$e−0.008t is always positive (hence there can be no stationary points and the proportion monotonically rises continuously). However in the limit, the gradient of successive tangents approaches the limiting value of $0$0.
We might wish for example to find the rate of discharge (the proportion discharged per day) when $t=100$t=100. Assuming the continuous model we can calculate $f'\left(100\right)=0.008e^{-0.008\times100}=0.003595$f′(100)=0.008e−0.008×100=0.003595.
In fact, without assuming a continuous model, we can compare this to the actual change in proportion between day $99$99 and day $100$100 given as:
$\frac{f\left(100\right)-f\left(99\right)}{100-99}$f(100)−f(99)100−99 | $=$= | $\frac{\left(1-e^{-0.008\times100}\right)-\left(1-e^{-0.008\times99}\right)}{100-99}$(1−e−0.008×100)−(1−e−0.008×99)100−99 |
$=$= | $\frac{\left(0.550671\right)-\left(0.547062\right)}{1}$(0.550671)−(0.547062)1 | |
$=$= | $0.003609$0.003609 | |
The two values ($0.003595$0.003595 as the gradient of the tangent at $t=100$t=100) and $0.003609$0.003609 (change in proportion between $t=99$t=99 and $t=100$t=100) essentially represent the rate of change of the proportion.
Here is the graph of the model, together with the tangent line drawn at $t=100$t=100. The gradient of this tangent line is approximately $0.003595$0.003595.
Tangents to the curve given by $y=\frac{1}{2}e^{-x^2}$y=12e−x2 drawn at its two inflection points intersect together at the point $A$A on the $y$y-axis. Find the area of the triangle formed by these two tangents and the $x$x-axis.
Since $y=\frac{1}{2}e^{-x^2}$y=12e−x2, we have $\frac{dy}{dx}=-xe^{-x^2}$dydx=−xe−x2 and for the second derivative:
$\frac{d^2y}{dx^2}$d2ydx2 | $=$= | $-2x\left(-xe^{-x^2}\right)+e^{-x^2}\left(-1\right)$−2x(−xe−x2)+e−x2(−1) |
$=$= | $e^{-2x^2}\left(2x^2-1\right)$e−2x2(2x2−1) | |
Setting $\frac{d^2y}{dx^2}=0$d2ydx2=0, we see that $2x^2-1=0$2x2−1=0 and so when solved, $x=\pm\frac{1}{\sqrt{2}}$x=±1√2. These must be the inflection points assumed by the question.
The tangent at $x=\frac{1}{\sqrt{2}}$x=1√2 has a gradient given by $m=-\left(\frac{1}{\sqrt{2}}\right)e^{-\left(\frac{1}{\sqrt{2}}\right)^2}=-\frac{1}{\sqrt{2e}}$m=−(1√2)e−(1√2)2=−1√2e and a point of contact of $\left(\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{e}}\right)$(1√2,12√e).
The equation of that tangent becomes:
$y-y_1$y−y1 | $=$= | $m\left(x-x_1\right)$m(x−x1) |
$y-\frac{1}{2\sqrt{e}}$y−12√e | $=$= | $-\frac{1}{\sqrt{2e}}\left(x-\frac{1}{\sqrt{2}}\right)$−1√2e(x−1√2) |
$\sqrt{e}y-\frac{1}{2}$√ey−12 | $=$= | $-\frac{1}{\sqrt{2}}x+\frac{1}{2}$−1√2x+12 |
$\sqrt{2}x+2\sqrt{e}y-2$√2x+2√ey−2 | $=$= | $0$0 |
The tangent intersects the $y$y axis when $x=0$x=0, so that $2\sqrt{e}y-2=0$2√ey−2=0 or that $y=\frac{1}{\sqrt{e}}$y=1√e. Thus the point $A$A has coordinates $\left(0,\frac{1}{\sqrt{e}}\right)$(0,1√e).
Because $y=\frac{1}{2}e^{-x^2}$y=12e−x2 is an even function, its curve is symmetrical about the y axis. This means that the tangent from the other inflection point will meet the $y$y axis at $\left(0,\frac{1}{\sqrt{e}}\right)$(0,1√e) also.
The tangents cut the $x$x axis when $y=0$y=0. For the tangent with the negative gradient, this means solving $\sqrt{2}x-2=0$√2x−2=0 from which $x=\sqrt{2}$x=√2.
From here, we can determine the base of the triangle as $2\sqrt{2}$2√2 and the height as $\frac{1}{\sqrt{e}}$1√e, so the area is given by $\frac{1}{2}\times b\times h=\frac{1}{2}\times2\sqrt{2}\times\frac{1}{\sqrt{e}}=\sqrt{\frac{2}{e}}$12×b×h=12×2√2×1√e=√2e.
The concentration, $x$x, of a certain medication in the bloodstream of a patient, $t$t hours after taking a dose, is given by $x=8te^{-4t}$x=8te−4t.
What is the concentration in the bloodstream when the patient initially takes the dose?
Solve for the time $t$t at which the patient has the greatest concentration of medication in his bloodstream.
Determine the maximum concentration of the medication. Round your answer to three decimal places.
According to the model, does the concentration ever reach $0$0 again?
Yes
No
Yes
No
Consider the function $f\left(x\right)=4e^{-x^2}$f(x)=4e−x2.
Determine $f'\left(x\right)$f′(x).
Determine the values of $x$x for which $f'\left(x\right)=0$f′(x)=0.
Determine the values of $x$x for which $f'\left(x\right)>0$f′(x)>0.
Determine the values of $x$x for which $f'\left(x\right)<0$f′(x)<0.
Determine the value of $\lim_{x\to\infty}f\left(x\right)$limx→∞f(x).
Determine the value of $\lim_{x\to-\infty}f\left(x\right)$limx→−∞f(x).
Which of the following is the graph of $f\left(x\right)$f(x)?
Consider the curve $f\left(x\right)=e^x+ex$f(x)=ex+ex.
By filling in the gaps, complete the proof showing that the tangent to the curve at the point $\left(1,2e\right)$(1,2e) passes through the origin.
The gradient function is given by | $f'\left(x\right)=\editable{}$f′(x)= | |
At $x=1$x=1, the gradient is | $f'\left(1\right)=\editable{}$f′(1)= | |
To find the equation of the tangent line with gradient $\editable{}$ and passing through point $\left(\editable{},\editable{}\right)$(,), we can use the point-gradient formula: |
$y-\editable{}=2e\left(x-\editable{}\right)$y−=2e(x−) | Equation 1 |
To determine whether the tangent line passes through the origin, we can substitute $x=\editable{}$x= into Equation 1, and check to see that the corresponding $y$y value is $\editable{}$ |
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When $x=\editable{}$x=: | $y-2e=\editable{}$y−2e= | |
$y=\editable{}$y= |
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