1. 2.1 The Tangent and Velocity Problems 1

2.  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. 2.1 The Tangent and Velocity Problems2

3. D EFINITION : Tangent line A tangent line to the function f(x) at the point, a, is a line that just touches the graph of the function at the point in question and is “parallel” (in some way) to the graph at that point. 32.1 The Tangent and Velocity Problems

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5.  Find the tangent line to at x =1. 52.1 The Tangent and Velocity Problems

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7.  Secant/Tangent applet Secant/Tangent applet 2.1 The Tangent and Velocity Problems7

8.  We can get a formula by finding the slope between P and Q using the “general” form of Pick some values of x getting closer and closer to x = 1, plug in and get some slopes. xM PQ x 2-60-2 1.5-50.5-3 1.1-4.20.9-3.8 1.01-4.020.99-3.98 1.001-4.0020.999-3.998 1.0001-4.00020.9999-3.9998 82.1 The Tangent and Velocity Problems

9.  Now, the equation of the line that goes through (a, f(a)) is given by Note: You are familiar with y - y 1 = m(x - x 1 )  Therefore, the equation of the tangent line to at x = 1 is Remember, a=1, f(a)=13, m=-4 92.1 The Tangent and Velocity Problems

10. 1. We wanted the tangent line to f(x) at a point x = a. First, we know that the point P=(a, f(a)) will be on the tangent line. Next, we’ll take a second point that is on the graph of the function, call it Q=(x, f(x)) and compute the slope of the line connecting P and Q as follows, 102.1 The Tangent and Velocity Problems

11. 2. We then take values of x that get closer and closer to (making sure to look at x’s on both sides of x = a) and use this list of values to estimate the slope of the tangent line, m. 3. The tangent line will then be, 112.1 The Tangent and Velocity Problems

12.  To compute the average rate of change of f(x) at all we need to do is to choose another point, say x, and then the average rate of change will be,  Then to estimate the instantaneous rate of change at all we need to do is to choose values of x getting closer and closer to a (don’t forget to chose them on both sides of a) and compute values of A.R.C. We can then estimate the instantaneous rate of change form that. 122.1 The Tangent and Velocity Problems

13.  Suppose that the amount of air in a balloon after t hours is given by Estimate the instantaneous rate of change of the volume after 5 hours. 132.1 The Tangent and Velocity Problems

14.  The first thing that we need to do is get a formula for the average rate of change of the volume. In this case this is, To estimate the instantaneous rate of change of the volume at t = 5 we just need to pick values of t that are getting closer and closer to t = 5. 142.1 The Tangent and Velocity Problems

15. xA.R.Cx 625.047.0 5.519.754.510.75 5.115.914.914.11 5.0115.09014.9914.9101 5.00115.0090014.99914.991001 5.000115.000900014.999914.99910001 Here is a table of values of t and the average rate of change for those values. So, from this table it looks like the average rate of change is approaching 15 and so we can estimate that the instantaneous rate of change is 15 at this point. 152.1 The Tangent and Velocity Problems

16.  Similar to rate of change.  To estimate the instantaneous velocity we would first compute the average velocity, 162.1 The Tangent and Velocity Problems

17.  Determine how far from x = a we want to move and then define our new point based on that decision. So, if we want to move a distance of h from x = a the new point would be x = a + h. We now get: 172.1 The Tangent and Velocity Problems

18.  A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after t minutes. When the data in the table are graphed, the slope of the tangent line represents the heart rate in beats per minute. The monitor estimates this value by calculating the slope of the secant line. Use the data to estimate the patient’s heart rate after 42 minutes using the secant line between the points with the given values of t. a) t = 36 and t = 42 b) t = 38 and t = 42 c) t = 40 and t = 42 d) t = 42 and t = 44 What are your conclusions? t (mins)3638404244 Heartbeats25302661280629483080 182.1 The Tangent and Velocity Problems

19.  If a rock is thrown upward on the planet Mars with a velocity of 10 m/s, its height in meters t seconds later is given by. a) Find the average velocity over the given time intervals: i. [1,2] ii. [1,1.5] iii. [1,1.1] iv. [1,1.01] v. [1,1.001] b) Estimate the instantaneous velocity when t = 1. 2.1 The Tangent and Velocity Problems19

20. 2.1 The Tangent and Velocity Problems20