Section 4.1 The Rate of Change of a Function
ΒΆCalculating and Interpreting the Slope of a Line.
Suppose we are given two points (x1,y1) and (x2,y2) on the line of a linear function y=f(x). Then the slope of the line is calculated by
We can interpret this equation by saying that the slope m measures the change in y per unit change in x. In other words, the slope m provides a measure of sensitivity .
Secant Line.
Secant is a Latin word meaning to cut, and in mathematics a secant line cuts an arbitrary curve described by y=f(x) through two points P and Q. The figure shows two such secant lines of the curve f to the right and to the left of the point P, respectively.
Definition 4.1. Slope of Secant Line β Average Rate of Change.
Suppose we are given two points (x1,y1) and (x2,y2) on the secant line of the curve described by the function y=f(x) as shown. Then the slope of the secant line is calculated by
Note that we may also be given the change in x directly as Ξx, i.e the two points are given as (x,f(x)) and (x+Ξx,f(x+Ξx)), and so
In the above figure, the value of Ξx must be negative since it is on the left side of x.
The slope of the secant line is also referred to as the average rate of change of f over the interval [x,x+Ξx].
The expression f(x+Ξx)βf(x)Ξx is referred to as the difference quotient.
Tangent Line.
Tangent is a Latin word meaning to touch, and in mathematics a tangent line touches an arbitrary curve described by y=f(x) at a point P but not any other points nearby as shown.
Example 4.2. Small Changes in x.
Consider y=f(x)=β625βx2 (the upper semicircle of radius 25 centered at the origin), and let's compute the changes of y resulting from small changes of x around x=7.
When \(x=7\text{,}\) we find that \(\ds y=\sqrt{625-49}=24\text{.}\) Suppose we want to know how much \(y\) changes when \(x\) increases a little, say to 7.1 or 7.01.
Let us look at the ratio \(\Delta y/\Delta x\) for our function \(\ds y=f(x)=\sqrt{625-x^2}\) when \(x\) changes from 7 to \(7.1\text{.}\) Here \(\Delta x=7.1-7=0.1\) is the change in \(x\text{,}\) and
Thus, \(\Delta y/\Delta x\approx -0.0294/0.1=-0.294\text{.}\) This means that \(y\) changes by less than one third the change in \(x\text{,}\) so apparently \(y\) is not very sensitive to changes in \(x\) at \(x=7\text{.}\) We say βapparentlyβ here because we don't really know what happens between 7 and value at \(7\text{.}\) This is not in fact the case for this particular function, but we don't yet know why.
Definition 4.3. Slope of Tangent LineβInstantaneous Rate of Change.
The slope of the tangent line to the graph of a function y=f(x) at the point P=(x,f(x)) is given by
provided this limit exists.
From Tangent Line Slope to Derivative.
Given a function f and a point x we can compute the derivative of f(x) at x as follows:
Form the difference quotient f(x+Ξx)βf(x)Ξx, which is the slope of a general secant line of the curve f throught the points P=(x,f(x)) and Q=(x+Ξx,f(x+Ξx)).
Take the limits as Ξx goes to zero: limΞxβ0f(x+Ξx)βf(x)Ξx, which is the slope of the tangent line of the curve f at the point P=(x,f(x)).
If this limit exists, then the derivative exists and is equal to this limit.
In other words,
provided the limit exists.
Subsection 4.1.1 Applications
ΒΆWe started this section by saying, βIt is often useful to know how sensitive the value of y is to small changes in x.β We have seen one purely mathematical example of this, involving the function f(x)=β625βx2. Here are some more applied examples. With careful measurement it might be possible to discover that the height of a dropped ball t seconds after it is released is h(t)=h0βkt2. (Here h0 is the initial height of the ball, when t=0, and k is some number determined by the experiment.) A natural question is then, βHow fast is the ball going at time t?β We can certainly get a pretty good idea with a little simple arithmetic.Example 4.4. Analyzing Velocity.
Suppose that the height h in metres of a dropped ball t seconds after it is released is given by
with h0=100 m and k=4.9. We will answer the question "How fast is the ball travelling at time t=2?" by exploring average speed near time t=2 and discussing the difference between speed and velocity.
We know that when \(t=2\) the height is \(100-4\cdot 4.9=80.4\) metres. A second later, at \(t=3\text{,}\) the height is \(100-9\cdot 4.9=55.9\) metres. The change in height during that second is \(55.9-80.4=-24.5\) metres. The negative sign means the height has decreased, as we expect for a falling ball, and the number 24.5 is the average speed of the ball during the time interval, in metres per second.
We might guess that 24.5 metres per second is not a terrible estimate of the speed at \(t=2\text{,}\) but certainly we can do better. At \(t=2.5\) the height is \(\ds 100-4.9(2.5)^2=69.375\) metres. During the half second from \(t=2\) to \(t=2.5\text{,}\) the change in height is \(69.375-80.4=-11.025\) metres giving an average speed of \(11.025/(1/2)=22.05\) metres per second. This should be a better estimate of the speed at \(t=2\text{.}\) So it's clear now how to get better and better approximations: compute average speeds over shorter and shorter time intervals. Between \(t=2\) and \(t=2.01\text{,}\) for example, the ball drops 0.19649 metres in one hundredth of a second, at an average speed of 19.649 metres per second.
We still might reasonably ask for the precise speed at \(t=2\) (the instantaneous speed) rather than just an approximation to it. For this, once again, we need a limit. Let's calculate the average speed during the time interval from \(t=2\) to \(t=2+\Delta t\) without specifying a particular value for \(\Delta t\text{.}\) The change in height during the time interval from \(t=2\) to \(t=2+\Delta t\) is
The average speed during this time interval is then
When \(\Delta t\) is very small, this is very close to 19.6. Indeed, \(\lim\limits_{\Delta t\to 0}\left(19.6+4.9\Delta t\right)=19.6\text{.}\) So the exact speed at \(t=2\) is 19.6 metres per second.
At this stage we need to make a distinction between speed and velocity. Velocity is signed speed, that is, speed with a direction indicated by a sign (positive or negative). Our algebra above actually told us that the instantaneous velocity of the ball at \(t=2\) is \(-19.6\) metres per second. The number 19.6 is the speed and the negative sign indicates that the motion is directed downwards (the direction of decreasing height).
In the language of the previous section, we might have started with \(\ds f(x)=100-4.9x^2\) and asked for the slope of the tangent line at \(x=2\text{.}\) We would have answered that question by computing
The algebra is the same. Thus, the velocity of the ball is the value of the derivative of a certain function, namely, of the function that gives the position of the ball.
Example 4.5. Demand for Sweaters.
A clothing manufacturer has determined that the weekly demand function of their sweaters is given by
where p is measured in dollars and q is measured in units of a thousand. Find the average rate of change in the unit price of a sweater if the quantity demanded is between 5000 and 6000 sweaters, between 5000 and 5100 sweaters, and between 5000 and 5010 sweaters. What is the instantaneous rate of change of the unit price when the quantity demanded is 5000 units?
The average rate of change of the unit price of a sweater if the quantity demanded is between \(q\) and \(q + \Delta q\) is
To find the average rate of change of the unit price of a sweater when the quantity demanded is between \(5000\) and \(6000\) sweaters (that is, over the interval \(\left[5,6\right]\)), we take \(q = 5\) and \(\Delta q = 1\text{,}\) obtaining
or -$\(11\) per \(1000\) sweaters. Similarly, taking \(\Delta q = 0.1\) and \(\Delta q = 0.01\) with \(q=5\text{,}\) we find that the average rates of change of the unit price when the quantities demanded are between \(5000\) and \(5100\) and between \(5000\) and \(5010\) are -$\(10.10\) and -$\(10.01\) per \(1000\) sweaters, respectively.
The instantaneous rate of change of the unit price of a sweater when the quantity demanded is \(q\) units is given by
In particular, the instantaneous rate of change when the quantity demanded is \(5000\) sweaters is
or -$\(10\) per \(1000\) sweaters.
Exercises for Section 4.1.
Exercise 4.1.1.
Draw the graph of the function \(\ds y=f(x)=\sqrt{169-x^2}\) between \(x=0\) and \(x=13\text{.}\) Find the slope \(\Delta y/\Delta x\) of the secant line between the points of the circle lying over (a) \(x=12\) and \(x=13\text{,}\) (b) \(x=12\) and \(x=12.1\text{,}\) (c) \(x=12\) and \(x=12.01\text{,}\) (d) \(x=12\) and \(x=12.001\text{.}\) Now use the geometry of tangent lines on a circle to find (e) the exact value of the derivative \(f'(12)\text{.}\) Your answers to (a)β(d) should be getting closer and closer to your answer to (e).
AnswerThe graph of \(y=f(x)=\sqrt{169-x^2}\) is a semi-circle of radius 13. We plot \(f(x)\) below on the interval \([0,13]\text{:}\)
We wish to estimate \(f'(12)\) first using secant lines. Therefore, we set up the difference quotient at \(x=12\text{:}\)
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The slope of the secant line between \(x=12\) and \(x=13\) (\(\Delta x = 1\)) is given by
\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{0-5}{1} = -5\text{.} \end{equation*} -
The slope of the secant line between \(x=12\) and \(x=12.1\) (\(\Delta x = 0.1\)) is given by
\begin{equation*} \frac{\Delta y}{\Delta x} \approx \frac{4.75289 - 5}{0.1} = -2.471\text{.} \end{equation*} -
The slope of the secant line between \(x=12\) and \(x=12.01\) (\(\Delta x = 0.01\)) is given by
\begin{equation*} \frac{\Delta y}{\Delta x} \approx \frac{4.97593 - 5}{0.01} = -2.407\text{.} \end{equation*} -
A property of circles is that the tangent line at any point on the circle is perpendicular the the line connecting that point to the centre of the circle (in this case, the origin):
The slope of the line connecting the point \((12,5)\) to the origin is\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{5-0}{12 - 0} = \frac{5}{12}\text{.} \end{equation*}Hence, the slope of the tangent line at \(x=12\) must be
\begin{equation*} -\frac{1}{\frac{5}{12}} = -\frac{12}{5} = 2.4\text{.} \end{equation*}We further notice that our answers to parts (a)-(d) are getting closer and closer to this value. Hence, \(f'(12) = 2.4\text{.}\)
Exercise 4.1.2.
Use geometry to find the derivative \(f'(x)\) of the function \(\ds f(x)=\sqrt{625-x^2}\) in the text for each of the following \(x\text{:}\) (a) 20, (b) 24, (c) \(-7\text{,}\) (d) \(-15\text{.}\) Draw a graph of the upper semicircle, and draw the tangent line at each of these four points.
AnswerThe graph of \(y=f(x)=\sqrt{625-x^2}\) is a semi-circle of radius 25. We plot \(f(x)\) below on the interval \([-25,25]\text{:}\)
We will use the fact that the tangent line at any point on the circle is perpendicular the the line connecting that point to the centre of the circle. For any point \((x, y) = (x, \sqrt{625-x^2})\text{,}\) the slope of the line connecting the origin to the point is given by
and so the slope of the tangent line at this point must be the reciprocal of this value:
-
At \(x=20\text{,}\) we have
\begin{equation*} f'(20) = - \frac{20}{\sqrt{625-20^2}} = -\frac{20}{15} = -\frac{4}{3}\text{.} \end{equation*} -
At \(x=24\text{,}\) we have
\begin{equation*} f'(24) = - \frac{24}{\sqrt{625-24^2}} = -\frac{24}{7}\text{.} \end{equation*} -
At \(x=-7\text{,}\) we have
\begin{equation*} f'(-7) = -\frac{-7}{\sqrt{625-(-7)^2}} = \frac{7}{24}\text{.} \end{equation*} -
At \(x=-15\text{,}\) we have
\begin{equation*} f'(-15) = -\frac{-15}{\sqrt{625-(-15)^2}} = \frac{15}{20} = \frac{3}{4}\text{.} \end{equation*}
We illustrate with the following graph.
Exercise 4.1.3.
Draw the graph of the function \(y=f(x)=1/x\) between \(x=1/2\) and \(x=4\text{.}\) Find the slope of the secant line between (a) \(x=3\) and \(x=3.1\text{,}\) (b) \(x=3\) and \(x=3.01\text{,}\) (c) \(x=3\) and \(x=3.001\text{.}\) Now use algebra to find a simple formula for the slope of the secant line between \((3,f(3))\) and \((3+\Delta x,f(3+\Delta x))\text{.}\) Determine what happens when \(\Delta x\) approaches 0. In your graph of \(y=1/x\text{,}\) draw the straight line through the point \((3,1/3)\) whose slope is this limiting value of the difference quotient as \(\Delta x\) approaches 0.
AnswerWe plot the graph of \(y = f(x) = 1/x\) over the interval \([0.5,4]\) below.
We first construct the difference quotient at \(x=3\text{:}\)
We use this difference quotient to compute the following.
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The slope of the secant line between \(x=3\) and \(x=3.1\) (\(\Delta x = 0.1\)) is
\begin{equation*} \frac{\Delta y}{\Delta x} = -\frac{1}{3(3+0.1)} \approx 0.1075\text{.} \end{equation*} -
The slope of the secant line between \(x=3\) and \(x=3.01\) (\(\Delta x = 0.01\)) is
\begin{equation*} \frac{\Delta y}{\Delta x} = -\frac{1}{3(3+0.01)} \approx 0.1107\text{.} \end{equation*} -
The slope of the secant line between \(x=3\) and \(x=3.001\) (\(\Delta x = 0.01\)) is
\begin{equation*} \frac{\Delta y}{\Delta x} = -\frac{1}{3(3+0.001)} \approx 0.1111\text{.} \end{equation*}
As \(\Delta x \to 0\text{,}\) we see that the slope of the secant line between \((3,f(3))\) and \((3+\Delta x, f(3+\Delta x)\) becomes
The equation of the line which passes through the point \((3,f(3)) = (3, 1/3)\) and has slope \(-1/9\) is
This is the equation of the tangent line at \(x=3\text{,}\) shown below.
Exercise 4.1.4.
Find an algebraic expression for the difference quotient \(\ds \bigl(f(1+\Delta x)-f(1)\bigr)/\Delta x\) when \(\ds f(x)=x^2-(1/x)\text{.}\) Simplify the expression as much as possible. Then determine what happens as \(\Delta x\) approaches 0. That value is \(f'(1)\text{.}\)
AnswerThe difference quotient for \(f(x) = x^2 - \frac{1}{x}\) is:
Therefore, at \(x=1\text{,}\) we have
Taking the limit as \(\Delta x \to 0\text{,}\) we find that
Exercise 4.1.5.
Draw the graph of \(\ds y=f(x)=x^3\) between \(x=0\) and \(x=1.5\text{.}\) Find the slope of the secant line between (a) \(x=1\) and \(x=1.1\text{,}\) (b) \(x=1\) and \(x=1.001\text{,}\) (c) \(x=1\) and \(x=1.00001\text{.}\) Then use algebra to find a simple formula for the slope of the secant line between \(1\) and \(1+\Delta x\text{.}\) (Use the expansion \(\ds (A+B)^3=A^3+3A^2B+3AB^2+B^3\text{.}\)) Determine what happens as \(\Delta x\) approaches 0, and in your graph of \(\ds y=x^3\) draw the straight line through the point \((1,1)\) whose slope is equal to the value you just found.
AnswerWe plot the graph of \(y=f(x) = x^3\) on the interval \([0,1.5]\) below.
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The slope of the secant line between \(x=1\) and \(x=1.1\) (\(\Delta x = 0.1\))is
\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{f(1.1)-f(1)}{0.1} = 3.31\text{.} \end{equation*} -
The slope of the secant line between \(x=1\) and \(x=1.001\) (\(\Delta x = 0.001\))is
\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{f(1.001)-f(1)}{0.1} = 3.003001\text{.} \end{equation*} -
The slope of the secant line between \(x=1\) and \(x=1.00001\) (\(\Delta x = 0.00001\))is
\begin{equation*} \frac{\Delta y}{\Delta x} = \frac{f(1.00001)-f(1)}{0.1} = 3.0000300001\text{.} \end{equation*}
We now simplify the difference quotient at \(x=1\text{:}\)
Therefore, as \(\Delta x \to 0\text{,}\)
The line with slope \(3\) which passes through the point \((1,f(1)) = (1,1)\) is
This is the tangent line to \(f\) at \(x=1\text{:}\)
Exercise 4.1.6.
Find an algebraic expression for the difference quotient \((f(x+\Delta x)-f(x))/\Delta x\) when \(f(x)=mx+b\text{.}\) Simplify the expression as much as possible. Then determine what happens as \(\Delta x\) approaches 0. That value is \(f'(x)\text{.}\)
AnswerLet \(f(x) = mx+b\text{.}\) The difference quotient is then
Therefore,
Exercise 4.1.7.
Sketch the unit circle. Discuss the behavior of the slope of the tangent line at various angles around the circle. Which trigonometric function gives the slope of the tangent line at an angle \(\theta\text{?}\) Why? Hint
We consider the following diagram of the unit circle and some \(\theta \in [0,\pi]\text{:}\)
The point \((a,b)\) is then given by \((\cos(\theta), \sin(\theta))\text{.}\) And so the slope of the line connecting the origin to this point is
Now, since the slope of the tangent line \(m\) at the point \((a,b)\) is perpendicular to the slope of the line connecting \((a,b)\) to the origin, we must have
This holds for any \(\theta \in [0, 2\pi]\text{.}\)
Exercise 4.1.8.
Sketch the parabola \(\ds y=x^2\text{.}\) For what values of \(x\) on the parabola is the slope of the tangent line positive? Negative? What do you notice about the graph at the point(s) where the sign of the slope changes from positive to negative and vice versa?
SolutionWe see that tangents lines which are tangent to the curve \(y = x^{2}\) at points \(x > 0\) have positive slope, and that tangent lines which are tangent to the curve at points \(x \lt 0\) have negative slope. We also see that the line tangent to \(x = 0\) has zero slope, and note that this corresponds to the vertex of the parabola: here, the minimum point on the graph.
Exercise 4.1.9.
An object is traveling in a straight line so that its position (that is, distance from some fixed point) is given by this table:
time (s) | 0 | 1 | 2 | 3 |
distance (m) | 0 | 10 | 25 | 60 |
Find the average speed of the object during the following time intervals: \([0,1]\text{,}\) \([0,2]\text{,}\) \([0,3]\text{,}\) \([1,2]\text{,}\) \([1,3]\text{,}\) \([2,3]\text{.}\) If you had to guess the speed at \(t=2\) just on the basis of these, what would you guess?
AnswerLet \(x\) be the distance travelled in metres and let \(t\) be the time in seconds. Then the average speed of the object over some time interval is \(\dfrac{\Delta x}{\Delta t}\text{.}\) Therefore:
For \(t \in [0,1]\text{,}\) the average speed is \(\dfrac{\Delta x}{\Delta t} = \dfrac{10-0}{1-0} = 10\) m/s.
For \(t \in [0,2]\text{,}\) the average speed is \(\dfrac{\Delta x}{\Delta t} = \dfrac{25-0}{2-0} = 12.5\) m/s.
For \(t \in [0,3]\text{,}\) the average speed is \(\dfrac{\Delta x}{\Delta t} = \dfrac{60-0}{3-0} = 20\) m/s.
For \(t \in [1,2]\text{,}\) the average speed is \(\dfrac{\Delta x}{\Delta t} = \dfrac{25-10}{2-1} = 15\) m/s.
For \(t \in [1,3]\text{,}\) the average speed is \(\dfrac{\Delta x}{\Delta t} = \dfrac{60-10}{3-1} = 25\) m/s.
For \(t \in [2,3]\text{,}\) the average speed is \(\dfrac{\Delta x}{\Delta t} = \dfrac{60-25}{3-2} = 35\) m/s.
We guess that the actual speed at \(t=2\) is between 15 and 35 m/s.
Exercise 4.1.10.
Let \(\ds y=f(t)=t^2\text{,}\) where \(t\) is the time in seconds and \(y\) is the distance in metres that an object falls on a certain airless planet. Draw a graph of this function between \(t=0\) and \(t=3\text{.}\) Make a table of the average speed of the falling object between (a) 2 sec and 3 sec, (b) 2 sec and 2.1 sec, (c) 2 sec and 2.01 sec, (d) 2 sec and 2.001 sec. Then use algebra to find a simple formula for the average speed between time \(2\) and time \(2+ \Delta t\text{.}\) (If you substitute \(\Delta t=1, 0.1, 0.01,0.001\) in this formula you should again get the answers to parts (a)β(d).) Next, in your formula for average speed (which should be in simplified form) determine what happens as \(\Delta t\) approaches zero. This is the instantaneous speed. Finally, in your graph of \(\ds y=t^2\) draw the straight line through the point \((2,4)\) whose slope is the instantaneous velocity you just computed; it should of course be the tangent line.
AnswerThe graph of \(y=f(t)=t^2\) on the interval \([0,3]\) is shown below, where \(y\) is the distance in metres of a falling object and \(t\) is the time in seconds.
The average speed of the falling object is given by \(\dfrac{\Delta y}{\Delta t}\text{.}\) Therefore:
Between \(t=2\) and \(t=3\text{,}\) the average speed is \(\dfrac{\Delta y}{\Delta t} = \dfrac{3^2-2^2}{3-2} = 5\) m/s.
Between \(t=2\) and \(t=2.1\text{,}\) the average speed is \(\dfrac{\Delta y}{\Delta t} = \dfrac{2.1^2-2^2}{2.1-2} = 4.1\) m/s.
Between \(t=2\) and \(t=2.01\text{,}\) the average speed is \(\dfrac{\Delta y}{\Delta t} = \dfrac{2.01^2-2^2}{3-2} = 4.01\) m/s.
Between \(t=2\) and \(t=2.001\text{,}\) the average speed is \(\dfrac{\Delta y}{\Delta t} = \dfrac{2.001^2-2^2}{3-2} = 4.001\) m/s.
Our formula for average speed between \(t=2\) and \(t=2+\Delta t\) can be simplified:
Therefore, the instantaneous speed is given by
So at \(t=2\text{,}\) the speed is 4 m/s (notice that our answers in (a)-(d) were approaching this value). The line which passes through the point \((2,4)\) and has slope \(4\) is given by
and is plotted below. This is the tangent line to \(y\) at \(t=2\text{.}\)
Exercise 4.1.11.
If an object is dropped from an 80-metre high window, its height \(y\) above the ground at time \(t\) seconds is given by the formula \(\ds y=f(t)=80-4.9t^2\text{.}\) (Here we are neglecting air resistance.) Find the average velocity of the falling object between (a) 1 sec and 1.1 sec, (b) 1 sec and 1.01 sec, (c) 1 sec and 1.001 sec. Now use algebra to find a simple formula for the average velocity of the falling object between 1 sec and \(1+\Delta t\) sec. Determine what happens to this average velocity as \(\Delta t\) approaches 0. That is the instantaneous velocity at time \(t=1\) second (it will be negative, because the object is falling).
AnswerLet \(y=f(t)=80-4.9t^2\) be the height in metres of a falling object above the ground at time \(t\) seconds.
The average velocity of the falling object is given by \(\dfrac{\Delta y}{\Delta t}\text{.}\) Notice that \(y(1) = 80-4.9 = 75.1\) m. Therefore:
Between \(t=1\) and \(t=1.1\text{,}\) the average speed is \(\dfrac{\Delta y}{\Delta t} = \dfrac{74.071-75.1}{0.1} = -10.29\) m/s.
Between \(t=1\) and \(t=1.01\text{,}\) the average speed is \(\dfrac{\Delta y}{\Delta t} = \dfrac{75.00151-75.1}{0.01} = -9.849\) m/s.
Between \(t=1\) and \(t=1.001\text{,}\) the average speed is \(\dfrac{\Delta y}{\Delta t} = \dfrac{75.0901951-75.1}{0.001} = -9.8049\) m/s.
The formula for average velocity between \(t=1\) and \(t=1+\Delta t\) can be simplified:
Therefore, the instantaneous velocity is given by
So at \(t=1\text{,}\) the velocity is -9.8 m/s (notice that our answers in (a)-(c) were approaching this value).
Exercise 4.1.12.
The following figure shows the devastating effect the opening of a new chain coffee shop had on a 3-generation rural cafΓ© in a small town. The revenue of the chain coffee shop at time \(t\) (in months) is given by \(f(t)\) million dollars, whereas the revenue of the rural cafΓ© at time \(t\) is given by \(g(t)\) million dollars. Answer the following questions by giving the value of \(t\) at which the specifying event took place.
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The revenue of the rural cafΓ© is decreasing at the slowest rate.
Answer\(t = 0\) -
The revenue of the rural cafΓ© is decreasing at the fastest rate.
Answer\(t = t_{3}\) -
The revenue of the chain coffe shop first overtakes that of the rural cafΓ©.
Answer\(t = t_{1}\) -
The revenue of the chain coffee shop is increasing at the fastest rate.
Answer\(t = t_{1}\)
Exercise 4.1.13.
The demand function for tires is given by
where \(p\) is measured in dollars and \(q\) is measured in units of a thousand.
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Find the average rate of change in the unit price of a tire if the quantity demanded is between \(5000\) and \(5050\) tires; between \(5000\) and \(5010\) tires.
AnswerSolution-$\(2.00\) per \(1000\) tires.Let's first form and simplify the difference quotient.
\begin{equation*} \begin{split} \dfrac{f(q+\Delta q) - f(q)}{\Delta q} \amp = \dfrac{\left(-0.1(q+\Delta q)^{2} - (q + \Delta q) + 125 \right) - \left(0.1q^{2} - q + 125\right)}{\Delta q} \\ \amp = \dfrac{-0.2q\Delta q-0.1\Delta q^{2} - \Delta q}{\Delta q}\\ \amp = -0.2q - 0.1\Delta q-1 \end{split} \end{equation*}Therefore, when \(q = 5\) for \(\Delta q = 0.05\) and \(0.01\text{,}\) we find that the difference quotient is equal to \(-2.005\) and \(-2.001\text{,}\) respectively. Therefore, when the quantity of tires demanded is between \(5000\) and \(5050\text{,}\) the unit price is decreasing by approximately $\(2.00\) per \(1000\) tires. And when the quantitity of tires demanded is between \(5000\) and \(5010\text{,}\) we again get that the unity price is decreasing by approximately $\(2.00\) per \(1000\) tires.
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What is the rate of change of the unit price if the quantity demanded is \(5000\text{?}\)
AnswerSolution-$\(2.00\) per \(1000\) tires.We find \(f'(x) = \lim\limits_{\Delta q \to 0}\left( -0.2q-0.1q\Delta q -1\right) = -0.2q - 1\text{.}\) Therefore, when the quantity demanded is \(5000\text{,}\) the unit price of the tires is changing by \(f'(5) = -2\text{.}\) That is, it is decreasing by $2.00 per \(1000\) tires.