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The Tangent Problems
The word tangent is derived from the Latin word tangens, which means “touching.”
Thus a tangent to a curve is a line that touches the curve.
In other words, a tangent line should have the same direction as the curve at the point of contact.
The Tangent Problems
For a circle we could simply follow Euclid and say that a tangent is a line that intersects the circle once and only once, as in Figure 1(a).
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The Tangent Problems
Figure 1(b) shows two lines l and t passing through a point P on a curve C.
The line l intersects C only once, but it certainly does not look like what we
think of as a tangent.
The line t, on the other hand, looks like a tangent but it intersects C twice.
Figure 1(b)
Example 1
Find an equation of the tangent line to the parabola y = x2 at the point P(1, 1).
Solution:
We will be able to find an equation of the tangent line t as soon as we know its slope m.
The difficulty is that we know only one point, P, on t, whereas we need two points to compute the slope.
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Example 1 – Solution
But observe that we can compute an approximation to m by choosing a nearby point Q(x, x2) on the parabola (as in
Figure 2) and computing the slope mPQ of the secant line PQ. [A secant line, from the Latin word secans,
meaning cutting, is a line that cuts (intersects) a curve more than once.]
cont’d
Figure 2
Example 1 – Solution
We choose x ≠ 1 so that Q ≠ P. Then
For instance, for the point Q(1.5, 2.25) we have
cont’d
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Example 1 – Solution
The tables below show the values of mPQ for several values of x close to 1.
The closer Q is to P, the closer x is to 1 and, it appears from the tables, the closer mPQ is to 2.
cont’d
Example 1 – Solution
This suggests that the slope of the tangent line t should be m = 2.
We say that the slope of the tangent line is the limit of the slopes of the secant lines, and we express this symbolically by writing
and
Assuming that the slope of the tangent line is indeed 2, we
cont’d
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Example 1 – Solution
Figure 3 illustrates the limiting process that occurs in this example.
Q approaches P from the right
Figure 3
cont’d
Example 1 – Solution
As Q approaches P along the parabola, the corresponding
Q approaches P from the left
Figure 3
cont’d
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The Velocity Problem
Example 3
Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground.
Find the velocity of the ball after 5 seconds.
Solution:
Through experiments carried out four centuries ago,
Galileo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. (This model for free fall neglects air
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Example 3 – Solution
If the distance fallen after t seconds is denoted by s(t) and measured in meters, then Galileo’s law is expressed by the equation
s(t) = 4.9t2
The difficulty in finding the velocity after 5 seconds is that we are dealing with a single instant of time (t = 5), so no time interval is involved.
cont’d
Example 3 – Solution
However, we can approximate the desired quantity by
computing the average velocity over the brief time interval of a tenth of a second from t = 5 to t = 5.1:
cont’d
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Example 3 – Solution
The following table shows the results of similar calculations of the average velocity over successively smaller time
periods.
cont’d
Example 3 – Solution
It appears that as we shorten the time period, the average velocity is becoming closer to 49 m/s.
The instantaneous velocity when t = 5 is defined to be the limiting value of these average velocities over shorter and shorter time periods that start at t = 5.
Thus it appears that the (instantaneous) velocity after 5 seconds is
cont’d
