dp studies y2. a. rates of change b. instantaneous rates of change c. the derivative function d....

Chapter 20Differential Calculus

DP Studies Y2

Contents:

A. Rates of changeB. Instantaneous rates of changeC. The derivative functionD. Rules of differentiationE. Equations of tangentsF. Normals to curves

OPENING PROBLEM

Valentino is riding his motorbike around a racetrack. A computer chip on his bike measures the distance Valentino has travelled as time goes on. This data is used to plot a graph of Valentino’s progress. Things to think about:

a. What is meant by a rate?b. What do we call the rate at which Valentino is travelling?c. What is the difference between an instantaneous rate and an average rate?d. How can we read a rate from a graph?e. How can we identify the fastest part of the racetrack?

A. Rates of Change

A rate is a comparison between two quantities of different kinds. For examples:

1. the speed at which a car is travelling in kmh-1 or ms-1.

2. the fuel efficiency of a car in kmL-1 or liters per 100 km travelled.

3. the scoring rate of a basketball player in points per game.

A. Rates of Change

Example 1:Josef typed 213 words in 3 minutes and made 6 errors, whereas Marie typed 260 words in 4 minutes and made 7 errors. Compare their performance using rates.

A. Rates of Change

Solution to example 1:

A. Rates of Change

If the graph which compares two quantities is a straight line, there is a constant rate of change in one quantity with respect to the other. This constant rate is the gradient of the straight line.

example:

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A. Rates of Change

If the graph is a curve, we can find the average rate of change between two points by finding the gradient of the chord or line segment between them. The average rate of change will vary depending on which two points are chosen, so it makes sense to talk about the average rate of change over a particular interval.

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A. Rates of Change

Example 2:The number of mice in a colony was recorded on a weekly basis.a. Estimate the average rate of increase in population for:

i. the period from week 3 to week 6ii. the seven week period.

b. What is the overall trend in the population growth over this period?

A. Rates of Change

b. The graph is increasing over the period by larger and larger amounts, so the population is increasing at an ever increasing rate.

B. Instantaneous Rate of Change

The instantaneous rate of change of a variable at a particularinstant is given by the gradient of the tangent to the graph atthat point.

Example:the graph alongside shows how a cyclist accelerates away

from an intersection. The average speed over the first 8 seconds is 100 m/8 sec = 12.5 ms-1.

Notice that the cyclist’s early speed is quite small, but it increases as time goes by.

Instantaneous rate of change example continue, To find the instantaneous speed at any time instant, for example t = 4, we draw the tangent to the graph at that time and find its gradient.

Finding the tangent gradient algebraically.Consider the curve y = x2 and the tangent at F(1, 1). Let the moving point M have x-coordinate 1 + h, where h ≠ 0. So, M is at (1 + h, (1 + h)2).

As point M gets closer and closer to point Fthe horizontal distance between the two points, h go to zero and the slope of the chord becomes 2.

Example 3:Use the algebraic method to find the gradient of the tangent to y = x2 at the point where x = 2.

Solution to example 3:

C. The Derivative Function

We can hence describe a gradient function which, for anygiven value of x, gives the gradient of the tangent at thatpoint. We call this gradient function the derived function orderivative function of the curve.

If we are given y in terms of x,we represent the derivative function by dy/dx.

If we are given the function f(x), we represent the derivative function by f‘(x).

C. The Derivative Function

The power rule of derivative

C. The Derivative Function

Example 4:

Solutions to example 4:

Example 5:

Solutions to example 5:

D. Rules of Differentiation

Differentiation is the process of finding a derivative or gradient function.

D. Rules of Differentiation

Example 6:

D. Rules of Differentiation

Solutions to example 6:a. f(x) = 5x3 + 6x2 – 3x + 2therefore, f’(x) = 5(3x2) + 6(2x) – 3(1) = 15x2 + 12x – 3

D. Rules of Differentiation

Example 7:

D. Rules of Differentiation

Solution to example 7:

D. Rules of Differentiation

Example 8:

D. Rules of Differentiation

Using the Ti-84 calculator, we can determine the derivative of a function by using the “CALCULATE” command on the graphing menu.

step 1: input the equation into y =step 2: graphstep 3: 2nd TRACE = CALCstep 4: choose “6”step 5: input x-value

D. Rules of Differentiation

y=x^3 + 3Graph2nd TRACE6-2dy/dx = 12.000001

So the gradient is 12 at x = -2 for y = x3 + 3

D. Rules of Differentiation

Example 9:

D. Rules of Differentiation

Solution to example 9:

D. Rules of Differentiation

Example 10:

D. Rules of Differentiation

Solution to example 10:

D. Rules of Differentiation

Example 11:The tangent to f(x) = 2x2 – ax + b at the point (2, 7) has a gradient of 3. Find a and b.

D. Rules of Differentiation

Solution to example 11:

E. Equations of Tangents

The equation of the tangent at point A(a, b) is

A(a, b)

E. Equations of Tangents

Example 12:Find the equation of the tangent to f(x) = x2 + 1 at the point

where x = 1.

E. Equations of Tangents

Solution to example 12:

E. Equations of Tangents

Example 13:Use technology to find the equation of the tangent to y = x3 – 7x + 3 at the point where x = 2.

E. Equations of Tangents

Solution to example 13:Using Ti-84,

y= y=function x^3 – 7x + 3graph2nd PRGMchoice “5”: Tangent(input the x-value 2read y = ax + b 5.000001x + -13.000002

So the equation of the tangent for y = x3 – 7x + 3 at the point x = 2 is y = 5x – 13

E. Equations of Tangents

Example 14: Consider the curve y = x3 – 4x2 – 6x + 8.

a. Find the equation of the tangent to this curve at the point where x = 0.

b. At what point does this tangent meet the curve again?

E. Equations of Tangents

Solution s to example 14

E. Equations of Tangents

In order to solve for part b, we graph both the original function and the tangent equation from part a.

y1 = x3 – 4x2 – 6x + 8

y2 = -6x + 8

Use the 2nd TRACE commandChoice 5enter x 3 and we should get x = 4, y = -16.

So the tangent meets the curve again at (4, -16)

F. Normals to Curves

A normal to a curve is a line which is perpendicular to the tangent at the point of contact.

Example1 5:Find the equation of the normal to f(x) = x2 – 4x + 3 at the point

where x = 4.

Solution to example 15:

Example 16:Find the coordinates of the point where the normal to y = x2 – 3

at (1, -2) meets the curve again.

Solution to example 16:

Using the calculator and the equation above, we can find the answer by graphing the two equations and find the intersect. The intersect occurs at (-1.5, -0.75)