Chapter 2.1 Lecture notes PDF

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Lecture notes for lecture 1 on chapter 2.1...

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2.1 Tangents and Velocities Discussion points: • Secant vs tangent lines • Finding the slope of secant and tangent lines • Meaning of the slope • Velocity Secant line:

Tangent line:

Question: How can we find the slope of secant lines and tangent lines?

Recall: To find the slope of a line we need to know at least two points on the line. Given P and Q, the line that goes through both points has a slope (m):

m=

Let P (2,4) and Q (5,25) be two points on the curve f(x) = x2. What is the slope of the secant line PQ?

m=

Problem: we only know one point (P) that the tangent line passes through.

Example: Let us take a look at some secant lines between the fixed point P and different points Q. The following table shows the slope of the secant lines PQ (correct to six decimal places) for the following values of Q (only mentioning xcoordinates here). (1) x = 3.9

mPQ = 6.222222

(5) x = 4.1 mPQ = 5.818182

(2) x = 3.99

mPQ = 6.020202

(6) x = 4.01 mPQ = 5.980198

(3) x = 3.999 mPQ = 6.002002

(7) x = 4.001 mPQ = 5.998002

(4) x = 3.9999 mPQ = 6.000200

(8) x = 4.0001 mPQ = 5.999800

Based on the table above, what would be a good guess for the value of the slope of the tangent line to the given curve at the given point P?

What can we observe based on the graph and the numerical data?

Review of lines and their slopes (Appendix B)

• Parallel lines • Perpendicular lines

• Equation of the line: Slope-intercept form

Point-slope form

Normal Line: the line that is perpendicular to the tangent line at the point of tangency. If P (2,4) is on the tangent line whose slope is 3, what is the equation of the tangent line? What is the equation of the normal line?

What does the slope represent? Slope of the secant line: the average rate of change of a function on an interval Slope of the tangent line: the instantaneous rate of change of a function at a point

Velocity Average Velocity =

𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒

Instantaneous Velocity is the limiting value of these average velocities over shorter and shorter time periods.

Example A rock is thrown upward on the planet Mars with a velocity of 15 m/s, its height (in meters) t seconds later is given by y = 15t − 1.86t2. (a) The average velocity over the given time intervals is given.

vavg =

s(t1 ) − s(t0 ) t1 − t0

(i) [1, 2] (ii) [1, 1.5]

9.42 m/s 10.35 m/s

(iii) [1, 1.1] (iv) [1, 1.01]

11.09 m/s 11.26 m/s

(v) [1, 1.001]

11.28 m/s

(b) Estimate the instantaneous velocity when t = 1....