AN INTRODUCTION TO CALCULUS THE TANGENT Calculus is divided into two main parts – differential calculus and integral calculus. These two branches of calculus were founded on two very familiar concepts respectively – tangents and areas, and although these two concepts seem completely unrelated, calculus has revealed that there is in fact, a close correlation between the two; a relationship that is defined by the Fundamental Theorem of Calculus. We'll explore this relationship when we start integral calculus. We begin this tutorial by exploring the 'founder' of differential calculus – the tangent. So, how do we define the term 'tangent'? Generally, a tangent could be said to be a line that touches a curve. But there is a serious flaw in that definition. If a tangent is indeed a line that merely touches a curve, then it means that a tangent could potentially touch the curve in any number of ways. It could pass through a curve, or intersect a curve twice, or possibly three times. The point is, this definition of a tangent is simply inaccurate. Euclid gives an acceptable definition of a tangent: a line that touches a curve once and only once . We will expand on this definition. Let's assume you have a point P(0.5, 2) lying on the curve y = 1/x, and we want to find the equation of the tangent line at point P. To solve this problem, we apply some basic coordinate geometry. To find the equation of a tangent, we use the slope-point form equation:
y – y1 = m(x – x1) Before we can move any further, we need to know what m (i.e. the slope) is. That's the first problem. This has led to another problem: we need two points to compute a slope, but we have only one point at the moment. The solution is to pick another point Q(x, 1/x) which has to be considerably close to P, and then compute the slope of the secant line PQ.
The red line is the tangent to point P, and the green line is the tangent to point Q. As you can see, we have picked x = 0.8 (which is close to 0.5). Thus, point Q will have coordinates (0.8, 1.25). So, the slope of PQ is given by MPQ
1.25 – 2
=
0.8 – 0.5
=
– 0.75 0.3
=
– 2.5
This means that when x = 0.8, then the slope of the secant line PQ equals -2.5. But let's not forget the original task; we're looking for the slope of the tangent line at P. What we need to do next is to pick other points that are much closer to P. The table below gives the values of MPQ for several values of x closer to 0.5:
x
Q(x, 1/x)
0.7 0.6 0.59 0.58 0.57 0.56 0.55 0.54 0.53 0.52 0.51 0.505 0.501 0.5001 0.50001
Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q
(0.7, 1.42857) (0.6, 1.66667) (0.59, 1.69491) (0.58, 1.72413) (0.57, 1.75439) (0.56, 1.78571) (0.55, 1.81818) (0.54, 1.85185) (0.53, 1.88679) (0.52, 1.92308) (0.51, 1.96078) (0.505, 1.98019) (0.501. 1.99601) (0.5001, 1.99960) (0.50001, 1.99996)
MPQ – – – – – – – – – – – – – – –
2.85715 3.33330 3.38989 3.44838 3.50871 3.57150 3.63640 3.70375 3.77367 3.84600 3.92200 3.96200 3.99000 4.00000 4.00000
On table 1, the values of x approach 0.5 from the right, and from the left on table 2. On both tables, observe the pattern: as x draws closer to 0.5 from both sides, the slope MPQ approaches –4. In other words, as Q approaches P, the slope of PQ approaches –4. Thus, we' ll assume at this point, that the slope is –4. So now, we can easily compute the equation of the tangent line at point P. We know that x1 = 0.5, y1 = 2, and m = –4. Thus, the equation becomes
y – 2 = –4(x – ½) y – 2 = –4x + 2 y = –4x + 2 + 2
TABLE 1
y = –4x + 4 y =
x 0.1 0.2 0.3 0.35 0.4 0.45 0.46 0.47 0.48 0.49 0.495 0.4955 0.499 0.4995
Q(x, 1/x) Q Q Q Q Q Q Q Q Q Q Q Q Q Q
(0.1, 10) (0.2, 5) (0.3, 3.33333) (0.35, 2.85714) (0.4, 2.5) (0.45, 2.22222) (0.46, 2.17391) (0.47, 2.12766) (0.48, 2.08333) (0.49, 2.04082) (0.495, 2.02020) (0.4955, 2.01816) (0.499, 2.00401) (0.49995, 2.00200)
4 – 4x
MPQ – – – – – – – – – – – – – –
20 10 6.66665 5.71427 5 4.4444 4.34775 4.25533 4.1665 4.082 4.04 4.03556 4.01 4.00
The idea of one quantity approaching a value as another related quantity approaches another value introduces the concept of a limit. This will be the subject of discussion in the next tutorial. Let's apply this concept to a simple, real situation: a ball thrown upwards.
TABLE 2
EXAMPLE 2 If a ball is thrown into the air with a velocity of 40ft/s, its height in feet after t seconds is given by y = 40t – 16t2. Find the instantaneous velocity when t = 2.
Solution We want to find the instantaneous velocity of the ball at EXACTLY 2 seconds. The motion of the ball is described explicitly by an equation. So, we'll graph the function to help give an idea of the situation:
We know that VELOCITY
=
Distance Traveled Time Elapsed
This implies that the velocity of the ball in 2 seconds is equal to the slope of the tangent line at P (see figure above). So, the task here is to find the slope of the tangent at P(2, 16). Like the first example above, we have the same problem: we need another point Q so that we'll be able to compute the slope of PQ. Of course, to obtain more accurate values of the slope, we use several points close to P. Here's another way of describing the situation: We need to compute the average velocities of the ball for several small time intervals. The instantaneous velocity at t = 2 will then be the limiting value of the average velocities. Hence we have the table below:
Time (s) 2.5 2.4 2.3 2.2 2.1 2,05 2.01 2.005
Height (ft) 0 3.84 7.36 10.56 13.44 14.76 15.76 15.88
TABLE 3
So, between 2 and 2.5 secs, AVERAGE VELOCITY
=
h(2.5) – h(2) 2.5 – 2
=
0 – 16 2.5 – 2
=
– 32ft/s
Between 2 and 2.4 secs, AVERAGE VELOCITY
=
h(2.4) – h(2) 2.4 – 2
=
3.84 – 16 2.4 – 2
= – 30.4ft/s
Note: the equation describing the motion of the ball is represented as h(t). The calculations above show the average velocities of the ball over small time intervals: 0.5 seconds and 0.4 seconds respectively. If we keep reducing the time intervals, the average velocity will definitely decrease, and we get the results on the following table:
t
h(t)
Q[t, h(t)]
2.5 2.4 2.3 2.2 2.1 2.05 2.01 2.005
0 3.84 7.36 10.56 13.44 14.76 15.76 15.88
Q Q Q Q Q Q Q Q
(2.5, 0) (2.4. 3.84) (2.3, 7.36) (2.2, 10.56) (2.1, 13.44) (2.05, 14.76) (2.01, 15.76 (2.005, 15.88)
MPQ –32 –30.4 –28.8 –27.2 –25.6 –24.8 –24.0 –24.0
From the calculations on the table 4, it is quite clear that the average velocities approach –24ft/s as the time intervals become smaller. This is the 'limit' of the average velocities as t approaches 2. This leads us to conclude that the instantaneous velocity of the ball when t = 2 equals –24ft/s.
TABLE 4 The first problem involved a function, and the second was a velocity problem. In both cases, we see that both situations were described by explicit equations. The next example shows how we apply the tangent concept to a function that is not described by an explicit equation, but by experimental data.
EXAMPLE 3 t (minutes)
Heartbeats
36
2530
the number of heartbeats after t minutes. When the data in the table are graphed, the
38
2661
slope of the tangent line represents the heart rate in beats per minute. The monitor
40
2806
42
2948
the patient's heart rate after 42 minutes using the secant lines between the points
44
3080
with the given values of t .
A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles
estimates this value by calculating the slope of a secant line. Use the data to estimate
Solution The first step is to plot the given table data and use them to sketch a curve:
From the graph, we see that the slope of the tangent line at P(42, 2948) represents the heart rate of the patient in beats per minute. First, we take point Q1 and compute the slope of the secant line PQ1:
Slope of secant line PQ1
2948 – 2530
=
42 – 36
=
69.7 beats/minute
=
71.75 beats/minute
=
71.00 beats/minute
=
66.00 beats/minute
We perform similar calculations for the other secant lines: Slope of secant line PQ2
=
Slope of secant line PQ3
=
Slope of secant line PQ4
=
2948 – 2661 42 – 38 2948 – 2806 42 – 40 2948 – 3080 42 – 44
From the results above, it doesn't seem as if the slopes are approaching any particular value. We therefore estimate the slope of the tangent line at P by averaging the slopes of the two closest secant lines: PQ3 and PQ4:
½ (71 + 66) = 68.5 This implies that the patient's heart rate after 42 minutes is 68.5 or approximately 69 beats/minute. Let's examine more problems on tangents and velocities. Please study each example carefully.
EXAMPLE 4 If an arrow is shot upward on the moon with a velocity of 58m/s, its height in meters after t seconds is given by h(t) = 58t – 0.83t2. ●
Find the average velocity over the given time intervals: (I) [1, 2]
●
(II) [1, 1.5]
(III) [1, 1.1]
(IV) [1, 1.01]
(V) [1, 1.001]
Find the instantaneous velocity after one second.
Solution The motion of the arrow is described by a function, which is graphed on the next page. From the question, we are looking for average velocities of the arrow over given time intervals. First, before we do anything else, we need to compute the height of the arrow after 2, 1.5, 1.1, 1.01, and 1.001 seconds:
t (seconds) h(t) (meters)
Q[t, h(t)]
2
112.680
Q1 (2, 112.680)
1
57.170
P (1, 57.170)
You'll notice that this problem is similar to Problem 2.
1.5
85.133
Q2 (1.5, 385.133)
1.1
62.796
Q3 (1.1, 62.796)
The next step is to compute the average velocities of the arrow over the given time intervals.
1.01
57.733
Q4 (1.01, 57.733)
1.001
57.226
Q5 (1.001, 57.226)
Therefore, from t = 1 to t = 2,
AVERAGE VELOCITY
=
h(1) – h(2)
=
1– 2
57.170 – 112.680 1–2
=
55.51 meters/sec
=
55.93 meters/sec
From t = 1 to t = 1.5,
AVERAGE VELOCITY
=
h(1) – h(1.5) 1 – 1.5
=
57.170 – 85.133 1 – 1.5
and so on. If perform similar average velocity calculations like the ones above, we end up with this table:
t (seconds) h(t) (meters) 2
112.680
1
Q[t, h(t)]
Average Velocity (m/s)
Q1 (2, 112.680)
55.51
57.170
P (1, 57.170)
--------
1.5
85.133
Q2 (1.5, 385.133)
55.93
1.1
62.796
Q3 (1.1, 62.796)
56.26
1.01
57.733
Q4 (1.01, 57.733)
56.30
1.001
57.226
Q5 (1.001, 57.226)
56.00
From the table above, it seems that the average velocities approach 56m/s. In other words, 56m/s is the limiting value of the average velocities. seconds:
Let's confirm our assumption by taking two more values of t closer to 1: 1.0001 and 1.00001
From t = 1, t = 1.0001,
AVERAGE VELOCITY
=
h(1) – h(1.0001) 1 – 1.0001
57.170 – 57.1756
=
1 – 1.0001
=
56.00 meters/sec
From t = 1, t = 1.00001, AVERAGE VELOCITY
=
h(1) – h(1.00001) 1 – 1.00001
=
57.170 – 57.17056 1 – 1.00001
=
56.00 meters/sec
Definitely, we can say that the limiting value of the average velocities is 56m/s. Since instantaneous velocity is equal to the limiting value of average velocities, it means that the instantaneous velocity of the arrow after one second is 56 m/s. The relationship between average velocity and tangents is given thus: If y = s(t) represents the motion of an object between time intervals [a, a+h], then we have the following:
SLOPE OF SECANT LINE PQ = AVERAGE VELOCITY
SLOPE OF TANGENT = INSTANTANEOUS VELOCITY
EXAMPLE 5 The point P(1, 0) lies on the curve y = sin(10π/x). (a) If Q is the point (x, sin(10π/x), find the slope of the secant line PQ (correct to four decimal places) for x = 2, 1.5, 1.4, 1.3, 1.2, 1.1, 0.5, 0.6, 0.7, 0.8, and 0.9. Do the slopes appear to be approaching a limit? (b) Use a graph of the curve to explain why the slopes of the secant lines in part (a) are not close to the tangent line at P. (c) By choosing appropriate secant lines, estimate the tangent line at P.
Solution (a). To solve problem (a), the first step is to evaluate the function y = sin(10π/x) for the given values of x. Thus, we have the table below:
x
y
The task here is to calculate the slope of the secant line PQ using the point Q(x,y). So,
2.0
0
When x = 2,
1.5
0.8660
1.4
-0.4339
1.3
-0.8230
1.2
0.8660
1.1
-0.2817
0.9
-0.3420
0.8
1.0
0.7
0.7818
0.6
0.8660
0.5
0
Slope of secant line PQ1
=
0
– 0
1 – 2
=
0
When x = 1.5, Slope of secant line PQ2
=
0 – 0.8860
=
1 – 1.5
1.732
When x = 1.4, Slope of secant line PQ3
=
0 – (– 0.4339) 1 – 1.4
=
- 1.0848
When x = 1.3, Slope of secant line PQ4
=
0 – (– 0.8230) 1 – 1.3
=
-2.7433
We perform similar calculations for the other values of x. In the end, we have this:
Q(x, y)
Slope of PQ
Q (2.0, 0)
0.0
Q (1.5, 0.8660)
1.7320
Q (1.4, -0.4339)
-1.0848
Q (1.3, -0.8230)
-2.7433
Q (1.2, 0.8660)
4.3300
Q (1.1, -0.2817)
-2.8170
Q (0.9, -0.3420)
3.4201
Q (0.8, 1.0)
-5.0
Q (0.7, 0.7818)
-2.6060
Q (0.6, 0.8660)
-2.1650
Q (0.5, 0)
0.0
(b). From the calculations and the table on the left, it is clear that the slopes of the secant lines PQ do NOT approach a limit. Instead, the slopes are fluctuating wildly, as the values of x approach 1 from both sides. So, what is really going on? Perhaps the graph on the following page will give an insight into this anomaly. If we graph the function in the viewing rectangle [-1, 3] by [-1.3, 1.3], we get this:
Weird, isn't it? Notice the massive oscillations. We can therefore say that the slopes of the secant lines do not appear to be approaching a limit because of the oscillations that occur between x = 0.5 and x = 2. If we draw tangent lines at the points Q(x, y) using the given values of x, we get this:
This completely explains the wild fluctuations in the slopes of the secant lines. (The colored lines represent the tangent lines of the function at the points Q(x, y)).
(c). To get a reasonable estimate of the slope of the tangent line at P, we need to choose an appropriate point, one that is very close to P. We therefore pick values like 1.01, 1.001, 1.0001 and perhaps 1.00001 (as close as possible). Using these values of x, we the following table::
Q(x, y)
Slope of PQ
Q (1.01, -0.3061)
-30.61
Q (1.001, -0.03137)
-31.37
Q (1.0001, -0.003141)
-31.41
Q (1.00001, -0.0003141)
-31.41
This table clearly shows that the slopes of the secant lines approach a value this time. If we take the two closest secant lines, we see that they have the same slope: -31.41. Thus, -31.41 is the limit of the slopes of the secant lines. We can therefore conclude that the slope of the tangent line P to the function y = sin(10π/x) is approximately -31.41
EXAMPLE 6 A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume V of water remaining in the tank (in gallons) after t minutes.
t (minutes)
5
10
15
20
25
30
V (gallons)
694
444
250
111
28
0
(a) If P is the point (15, 250) on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graph with t = 5, 10, 20, 25, and 30. (b) Estimate the slope of the tangent line at P by averaging the slopes of the two secant lines. (c) Use a graph of the function to estimate the slope of the tangent line at P. (This slope represents the rate at which the water is flowing from the tank after 15 minutes.)
Solution (a). This example involves a function with two quantities: time and volume; quantities that are related by tabular data, not an equation. We can use the data to sketch a curve. We want to find the slopes of the secant lines PQ, where P has coordinates (15, 250), and Q has coordinates (t, V). Therefore, the slope of the secant line PQ will be given by Slope of secant line PQ
=
V – 250 t – 15
So, when t = 5, Slope of secant line PQ1
=
694 – 250 5 – 15
=
- 44.4 gallons/minute
=
- 38.8 gallons/minute
When t = 10, Slope of secant line PQ2
=
444 – 250 10 – 15
When t = 20, Slope of secant line PQ3
111 – 250
=
20 – 15
=
- 27.8 gallons/minute
=
- 22.2 gallons/minute
=
- 16.7 gallons/minute
When t = 25, Slope of secant line PQ4
=
28 – 250 25 – 15
When t = 30, Slope of secant line PQ5
=
0 – 250 30 – 15
Notice that all the slopes are negative. This indicates that more water is drained from the tank as time elapses.
`
(b). To compute the average slope at P, we use the closest points to P. In this case, the closest points are Q2 and Q3. The corresponding slopes PQ2 and PQ3 are – 38.8 and – 27.8 respectively. Thus, the average slope is Average Slope of secant line PQ
= =
– 38.8 + (– 27.8) 2
=
– 38.8 – 27.8 2
- 33.3 gallons/minute
With this method, we estimate that water is drained from the tank at a rate of roughly 33.3 gallons per minute.
(c). In (b), we averaged the two closest slopes to obtain an estimate of the slope at P. We obtained -33.3 gallons/minute. There's another method we can use to estimate the slope:
X
Y
Z
Observe the graph: We draw an approximate tangent to P, and project a triangle XYZ. We measure the sides XY and YZ. Thus, an estimate of the slope is given by:
–
|XY| |YZ|
Approximately, we find that |XY| = 300, and |YZ| = 9.01. Therefore,
–
|XY| |YZ|
≈
–
300 9.01
≈
- 33.296
Using two methods, we obtained -33.3 and -32.296, which both amount to roughly -33.3 in the long run. Thus, we say that, after 15 minutes, water is flowing from the tank (or the tank is losing water) at the rate of 33.3 gallons per minute. Try the following exercises.
EXERCISE 1 The position of a car is given by the values in the table:
(a)
t (seconds)
0
1
2
3
4
5
s (feet)
0
10
32
70
119
178
Find the average velocity for the time period beginning when t = 2 and lasting (i) 3 seconds
(b)
(ii) 2 seconds
(iii) 1 second
Use the graph of s as a function of t to estimate the instantaneous velocity when t = 2.
EXERCISE 2 The point P(4, 2) lies on the curve y = √x. (a) If Q is the point (x, √x), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for values of x close to 4. (b) Use the results of part (a) to guess the value of the slope of the tangent line to the curve at P(4, 2). (c) Use the slope obtained from part (b) to find an equation of the tangent line to the curve at P(4, 2)
EXERCISE 3 The displacement (in feet) of a certain particle moving in a straight line is given by s = t3/6, where t is measured in seconds. (a) Find the average over the following time periods: (i) [1, 3]
(ii) [1, 2]
(iii) [1, 1.5]
(iv) [1, 1.1]
(b) Find the instantaneous velocity when t = 1
In the following section, we turn our full attention to limits, and how we deal with them.
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