Exercise 3.1:

Draw the graph of the function y = between x = 0 and x = 13. Find the slope of the chord between the points of the circle lying over (a) x = 12 and x = 13, (b) x = 12 and x = 12.1, (c) x = 12 and x = 12.01, (d) x = 12 and x = 12.001. Now use the geometry of tangent lines on a circle to find (e) the exact value of the derivative f'(12). Your answers to (a)--(d) should be getting closer and closer to your answer to (e).

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Exercise 3.2:

Use geometry to find the derivative f'(x) of the function in the text for each of the following x: (a) 20, (b) 24, (c) -7, (d) -15. Draw a graph of the upper semicircle, and draw the tangent line at each of these four points.

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Exercise 3.3:

Let y = , where t is the time in seconds and y is the distance in meters that an object falls on a certain airless planet. Draw a graph of this function between t = 0 and t = 3. Make a table of the average velocity of the falling object between (a) 2 sec and 3 sec, (b) 2 sec and 2.1 sec, (c) 2 sec and 2.01 sec, (d) 2 sec and 2.001 sec. Then use algebra to find a simple formula for the average velocity between time 2 and time . (If you substitute , , , in this formula you should again get the answers to parts (a)--(d).) Next, in your formula for average velocity (which should be in simplified form) determine what happens as approaches zero. This is the instantaneous velocity. Finally, in your graph of draw the straight line through the point (2,4) whose slope is the instantaneous velocity you just computed.

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Exercise 3.4:

If an object is dropped from an 80-meter high window, its height y above the ground at time t sec is given by the formula y = . (Here we are neglecting air resistance; the graph of this function was shown at the beginning of the first section.) Find the average velocity of the falling object between (a) 1 sec and 1.1 sec, (b) 1 sec and 1.01 sec, (c) 1 sec and 1.001 sec. Now use algebra to find a simple formula for the average velocity of the falling object between 1 sec and . Determine what happens to this average velocity as approaches 0. That is the instantaneous velocity at time t = 1 sec (it'll be negative, because the object is falling).
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Exercise 3.5:

Draw the graph of the function y = between and x = 4. Find the slope of the chord between (a) x = 3 and x = 3.1, (b) x = 3 and x = 3.01, (c) x = 3 and x = 3.001. Now use algebra to find a simple formula for the slope of the chord between ( ) and ( ). Determine what happens when approaches 0. In your graph of , draw the straight line through the point ( ) whose slope is this limiting value of the difference quotient as approaches 0.

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Exercise 3.6:

Find an algebraic expression for the difference quotient when . Simplify the expression as much as possible. Then determine what happens as approaches 0. That value is f'(1).

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Exercise 3.7:

Draw the graph of y = between x = 0 and x = 1.5. Find the slope of the chord between (a) x = 1 and x = 1.1, (b) x = 1 and x = 1.001, (c) x = 1 and x = 1.00001. Then use algebra to find a simple formula for the slope of the chord between 1 and . (Use the expansion: ) Determine the limit as approaches 0, and in your graph of draw the straight line through the point (1,1) whose slope is equal to the limit you just found.

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