[Fig. 2.5]
Chapter 1 started with straight line graphs. The velocity was constant (at least piece-wise). The distance function was linear. Now we are facing polynomials like x³ − 2 or x⁴ − x² + 3, with other functions to come soon. Their graphs are definitely curved. Most functions are not close to linear—except if you focus all your attention near a single point. That is what we will do.
Over a very short range a curve looks straight. Look through a microscope, or zoom in with a computer, and there is no doubt. The graph of distance versus time becomes nearly linear. Its slope is the velocity at that moment. We want to find the line that the graph stays closest to—the "tangent line"—before it curves away.
The tangent line is easy to describe. We are at a particular point on the graph of y = ƒ(x). At that point x equals a and y equals ƒ(a) and the slope equals ƒ'(a). The tangent line goes through that point x = a, y = ƒ(a) with that slope m = ƒ'(a). Figure 2.5 shows the line more clearly than any equation, but we have to turn the geometry into algebra. We need the equation of the line.
EXAMPLE 1 Suppose y = x⁴ − x² + 3. At the point x = a = 1, the height is y = ƒ(a) = 3. The slope is dy/dx = 4x³ − 2x. At x = 1 the slope is 4 − 2 = 2. That is ƒ'(a):
The numbers x = 1, y = 3, dy/dx = 2 determine the tangent line.
The equation of the tangent line is y − 3 = 2(x − 1), and this section explains why.
[Fig. 2.5]
A straight line is determined by two conditions. We know the line if we know two of its points. (We still have to write down the equation.) Also, if we know one point and the slope, the line is set. That is the situation for the tangent line, which has a known slope at a known point:
I will take those one at a time-first y = mx + b, then m, then b.
1. The graph of y = mx + b is not curved. How do we know? For the specific example y = 2x + 1, take two points whose coordinates x, y satisfy the equation:
x=0, y= 1 and x=4, y=9 both satisfy y= 2x + 1.
Those points (0, 1) and (4,9)lie on the graph. The point halfway between has x = 2 and y = 5. That point also satisfies y = 2x + 1. The halfway point is on the graph. If we subdivide again, the midpoint between (0, 1) and (2, 5) is (1, 3). This also has y = 2x + 1. The graph contains all halfway points and must be straight.
2. What is the correct slope m for the tangent line? In our example it is m = ƒ'(a) = 2. The curve and its tangent line have the same slope at the crucial point: dy/dx = 2.
Allow me to say in another way why the line y = mx + b has slope m. At x = 0 its height is y = b. At x = 1 its height is y = m + b. The graph has gone one unit across (0 to 1) and m units up (b to m + b). The whole idea is
slope = distance up ⁄ distance across = m ⁄ 1. (1)
Each unit across means m units up, to 2m + b or 3m + b. A straight line keeps a constant slope, whereas the slope of y = x⁴ − x² + 3 equals 2 only at x = 1.
3. Finally we decide on b. The tangent line y = 2x + b must go through x = 1, y = 3. Therefore b = 1. With letters instead of numbers, y = mx + b leads to ƒ(a) = ma + b. So we know b:
2E The equation of the tangent line has b = ƒ(a) − ma: y= mx + ƒ(a) − ma or y − ƒ(a) = m(x − a). (2)
That last form is the best. You see immediately what happens at x = a. The factor x − a is zero. Therefore y = ƒ(a) as required. This is the point-slope form of the equation, and we use it constantly:
y − 3 = 2(x − 1) or (y &minus 3) ⁄ (x − 1) = distance up ⁄ distance across = slope 2.
EXAMPLE 2 The curve y = x³ − 2 goes through y = 6 when x = 2. At that point dy/dx = 3x² = 12. The point-slope equation of the tangent line uses 2 and 6 and 12:
y − 6 = 12(x − 2), which is also y= 12x − 18.
[Fig. 2.6]
There is another important line. It is perpendicular to the tangent line and perpen-dicular to the curve. This is the normal line in Figure 2.6. Its new feature is its slope. When the tangent line has slope m, the normal line has slope -llm. (Rule: Slopes of perpendicular lines multiply to give -1.) Example 2 has m = 12, so the normal line has slope -1/12:
tangent line: y − 6 = 12(x − 2) normal line: y − 6 = -(1/12)(x − 2).
Light rays travel in the normal direction. So do brush fires—they move perpendicular to the fire line. Use the point-slope form! The tangent is y = 12x − 18, the normal is not y = -(1/12)x &minus 18.
EXAMPLE 3 You are on a roller-coaster whose track follows y = x² + 4. You see a friend at (0,O)and want to get there quickly. Where do you step off?
Solution Your path will be the tangent line (at high speed). The problem is to choose x = a so the tangent line passes through x = 0, y = 0. When you step off at x = a,
the height is y = a² + 4 and the slope is 2a
the equation of the tangent line is y − (a² + 4) = 2a(x − a)
this line goes through (0,0) if -(a² + 4)= -2a² or a = ±2.
The same problem is solved by spacecraft controllers and baseball pitchers. Releasing a ball at the right time to hit a target 60 feet away is an amazing display of calculus. Quarterbacks with a moving target should read Chapter 4 on related rates.
Here is a better example than a roller-coaster. Stopping at a red light wastes gas. It is smarter to slow down early, and then accelerate. When a car is waiting in front of you, the timing needs calculus:
EXAMPLE 4 How much must you slow down when a red light is 72 meters away? In 4 seconds it will be green. The waiting car will accelerate at 3 meters/sec². You cannot pass the car.
Strategy Slow down immediately to the speed V at which you will just catch that car. (If you wait and brake later, your speed will have to go below V.)At the catch-up time T, the cars have the same speed and same distance. Two conditions, so the distance functions in Figure 2.6d are tangent.
Solution At time T, the other car's speed is 3(T − 4). That shows the delay of 4 seconds. Speeds are equal when 3(T − 4)= V or T = (1/3)V + 4. Now require equal distances. Your distance is V times T. The other car's distance is 72 + ½at²:
72 + ½·3(T − 4)² = VT becomes 72 + ½·(1/3)V² = V(3V+4).
The solution is V = 12 meters/second. This is 43 km/hr or 27 miles per hour.
Without the other car, you only slow down to V = 72/4 = 18 meters/second. As the light turns green, you go through at 65 km/hr or 40 miles per hour. Try it.
Instead of the tangent line through one point, consider the secant line through two points. For the tangent line the points came together. Now spread them apart. The point-slope form of a linear equation is replaced by the two-point form.
The equation of the curve is still y = ƒ(x). The first point remains at x = a, y = ƒ(a). The other point is at x = c, y = ƒ(c). The secant line goes between them. and we want its equation. This time we don't start with the slope—but m is easy to find.
EXAMPLE 5 The curve y = x³ − 2 goes through x = 2, y = 6. It also goes through x = 3, y = 25. The slope between those points is
m = change in y ⁄ change in x = (25-6) ⁄ (3-2) = 19.
The point-slope form (at the first point) is y − 6 = 19(x − 2). This line automatically goes through the second point (3,25). Check: 25-6 equals 19(3-2). The secant has the right slope 19 to reach the second point. It is the average slope Δy/Δx.
A look ahead The second point is going to approach the first point. The secant slope Δy/Δx will approach the tangent slope dy/dx. We discover the derivative (in the limit). That is the main point now—but not forever.
Soon you will be fast at derivatives. The exact dyldx will be much easier than Δy/Δx. The situation is turned around as soon as you know that x⁹ has slope 9x⁸. Near x = 1, the distance up is about 9 times the distance across. To find Δy = 1.0019-19,just multiply Δx = .001 by 9. The quick approximation is .009, the calculator gives Δy = .009036. It is easier to follow the tangent line than the curve.
Come back to the secant line, and change numbers to letters. What line connects x = a, y = ƒ(a) to x = c, y = ƒ(c)? A mathematician puts formulas ahead of numbers, and reasoning ahead of formulas, and ideas ahead of reasoning:
2F The two-point form uses the slope between the points: secant line: y − ƒ(a) = {(ƒ(c) &minus ƒ(a)) ⁄ (c − a)}(x − a). (3)
At x = a the right side is zero. So y = ƒ(a) on the left side. At x = cthe right side has two factors c − a. They cancel to leave y = ƒ(c). With equation (2) for the tangent line and equation (3) for the secant line, we are ready for the moment of truth.
What comes now is pretty basic. It matches what we did with velocities:
average velocity = Δ distance ⁄ Δ time = (ƒ(t+Δt) − ƒ(t)) ⁄ Δt.
The limit is dƒ/dt. We now do exactly the same thing with slopes. The secant tine turns into the tangent line as capproaches a:
slope of secant line: Δƒ/Δx = (ƒ(c) − ƒ(a)) ⁄ (c − a)
slope of tangent line: dƒ/dx = limit of Δƒ/Δx.
There stands the fundamental idea of differential calculus! You have to imagine more secant lines than I can draw in Figure 2.7, as c comes close to a. Everybody recognizes c − a as Δx. Do you recognize ƒ(c) − ƒ(a) as ƒ(x + Δx) − ƒ(x)? It is Δƒ, the change in height. All lines go through x = a, y = ƒ(a). Their limit is the tangent line.
[Fig. 2.7]
Intuitively, the limit is pretty clear. The two points come together, and the tangent line touches the curve at one point. (It could touch again at faraway points.) Mathematically this limit can be tricky—it takes us from algebra to calculus. Algebra stays away from 0/0, but calculus gets as close as it can.
The new limit for dƒ/dx looks different, but it is the same as before:
ƒ'(a) = limc→a (ƒ(c) − ƒ(a)) ⁄ (c − a).
EXAMPLE 6 Find the secant lines and tangent line for y = ƒ(x) = sin x at x = 0.
The starting point is x = 0, y = sin 0. This is the origin (0,0). The ratio of distance up to distance across is (sin c)/c:
secant equation y = (sinc x ⁄ c)x tangent equation y = 1x.
As c approaches zero, the secant line becomes the tangent line. The limit of (sin c)/c is not 0/0, which is meaningless, but 1, which is dy/dx.
EXAMPLE 7 The gold you own will be worth √t million dollars in t years. When does the rate of increase drop to 10% of the current value, so you should sell the gold and buy a bond? At t = 25, how far does that put you ahead of √t = 5?
Solution The rate of increase is the derivative of √t,which is 1/2√t. That is 10% of the current value √t when 1/2√t = √t/10. Therefore 2t = 10 or t = 5. At that time you sell the gold, leave the curve, and go onto the tangent line:
y − √5 = √5/10 (t &minus 5) becomes y − √5 = 2√5 at t = 25.
With straight interest on the bond, not compounded, you have reached y = 3√5 = 6.7 million dollars. The gold is worth a measly five million.
