2.7 Derivatives and Rates of Change Tangents Let us …rst of all revise some of the topics we discussed in “Section 2.1. Tangent and Velocity Problems”. Secant Line This is a line passing through any two points on a curve y = f (x). Let P (a; f (a)) and Q(x; f (x)) be the two points.

The slope of the secant line is f (x) f (a) : x a Recall that as point Q become closer and closer to P , the secant line approaches the tangent line at P . mP Q =

Tangent Line This is the line that intersects the curve at only one point. Let P (a; f (a)) be the point.

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We have the following arguments. 1. As point Q approaches P , i.e. as x approaches a, or x ! a, 2. the secant line P Q approaches the tangent line at P , and then 3. the slope of the secant line P Q approaches the slope of the tangent line at P . 4. We know that the slope of the secant line P Q is mP Q =

f (x) x

f (a) . a

5. Therefore, the slope of the tangent line at P (a; f (a)) is m = lim

x!a

f (x) x

f (a) : a

Example 1 Find the equation of the tangent line to the parabola y = x2 at the point (1; 1).

Another way to calculate the slope of the tangent line at x = a If we know that point Q is h distance horizontally away from point P (a; f (a)), we can write point Q as Q (a + h; f (a + h)).

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Then we have 1. The slope of the secant line can be rewritten as f (a + h) f (a) f (a + h) mP Q = = (a + h) a h

f (a)

2. The secant line approaches the tangent line at P as Q approaches P , or as a + h ! a, i.e. h ! 0. 3. Therefore, the slope of the tangent line can be calculated from the slope of the secant as f (a + h) f (a) m = lim h!0 h We can then de…ne the slope of the tangent line as following De…nition 1 (Tangent Line) The tangent line to the curve y = f (x) at the point P (a; f (a)), is the line through point P with slope f (a + h) f (x) f (a) or m = lim x!a h!0 x a h provided that this limit exists. m = lim

f (a)

Example 2 Find an equation of the tangent line to the hyperbola y = point (3; 1).

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1 Example 3 Find the slope of the tangent line to the curve y = p at the x point where x = a.

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Velocities Recall in Section 2.1, we plotted the secant line on the graph of displacement travelled vs. the time used. In the graph given below, s(t) is the position of an object at time t, and it is called position function.

The average velocity of an object having travelled from s(t) to s(t + h) in time h is the slope of the secant line P Q, displacement time s(t + h) s(t) = (t + h) t s(t + h) s(t) = h

Average velocity =

Then the instantaneous velocity or the velocity v(t) at time t, is de…ned to be the limit of the average velocity. De…nition 2 (Instantaneous Velocity) The instantaneous velocity v(a) of the position function s(t) at time t = a is given by s(a + h) h!0 h

v(a) = lim

s(a)

or v(a) = lim t!a

s(t) t

s(a) a

provided the limit exists. In Stewart’s textbook on page 145, the position function is denoted by f (t).

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Example 4 Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground. Use s(t) = 4:9t2 to be the equation of motion of the ball. (1) What is the velocity of the ball after 5 seconds? (2) How fast is the ball travelling when the ball hits the ground?

Derivatives De…nition 3 The derivative of a function f at a number a, denoted by f 0 (a), is f (a + h) f (a) f (x) f (a) or f 0 (a) = lim f 0 (a) = lim x!a h!0 x a h provided the limit exists. Example 5 Find the derivative of the function f (x) = x2 number a using the de…nition of derivatives.

8x + 9 at the

Equation of the tangent line The derivative of f at a, f 0 (a), is the slope of the tangent line to y = f (x) at (a; f (a)). Therefore, the point-slope form of the equation of the tangent line to the curve y = f (x) at the point (a; f (a)) is y

f (a) = f 0 (a)(x 5

a):

Example 6 Find the equation of the tangent line to the parabola f (x) = x2 8x + 9 at the point (3; 6).

Rates of Change Let us look at the graph of y = f (x) where point P (x1 ; f (x1 )) changes to point Q(x2 ; f (x2 )) when there is a change in the value of x.

When x changes in value from x1 to x2 , the change in x is 4x = x2

x1

and the corresponding change in y is 4y = f (x2 )

f (x1 ):

The slope of the secant line P Q is the quotient mP Q =

Rise 4y f (x2 ) = = Run 4x x2

f (x1 ) ; x1

which is called the average rate of change of y with respect to x over the interval [x1 ; x2 ]: Then the instantaneous rate of change of y with respect to x at x = x1 is the slope of the tangent line to the curve y = f (x) at P . 6

De…nition 4 The instantaneous rate of change of y with respect to x at x = x1 is given by f (x2 ) 4y = lim x2 !x1 4x!0 4x x2

instantaneous rate of change = lim

f (x1 ) x1

provided the limit exists. We can use 4x in the same way as we use h. After all, they mean the same thing. Stewart (Section 2.7): The derivative f (a) is the instantaneous rate of change of y = f (x) with respect to x when x = a. What does this mean? When the absolute value of the derivative is large, the curve is steep and the y-values change rapidly. When the absolute value of the derivative is small, the curve is almost ‡at and the y-values change slowly. One way to think about it: Velocity is the derivative of the position function. When the absolute value of your velocity is large i.e. you drive at breakneck speed, you change position rapidly. When the absolute value of your velocity is small i.e. you drive at snail’s pace, you change position slowly. Example 7 The position of a particle is given by the equation of motion 1 , where t is measured in seconds and s(t) is measured in metres. s(t) = 1+t Find the velocity and the speed of the particle after 2 seconds have elapsed.

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