How Secant Lines become Tangent Lines

Adrienne, Samantha, Danielle, and Eugene, trying to raise Walkathon money, build a 196 foot high-dive platform in the middle of the plaza. After charging admission for prime bleacher seats in the plaza, they then “persuade” Mr Murphy to be the first high diver. Abby, Nica, and Hamidou observe Mr Murphy’s “dive” closely enough to form an equation for his height. They find the equation to be given by: …and the graph is given by…

s ( t ) = H e i g h t o f f o f t h e g r o u n d ( i n f e e t ) t = Time in seconds

What are the coordinates for this point? (0, 196) because at time = 0, Mr. Murphy is at the top of the 196 foot high platform. At what time did Mr. Murphy land? (3.5, 0) which you can find by setting s(t) = 0 What is Mr. Murphy’s average velocity during his 3.5 second plunge? = −56 feet/sec Why is the velocity negative? Because the motion is downward. s(t) = Height off of the ground(in feet) t = Time in seconds

Since you are all experts at algebra… =  56 feet/sec How can you represent the average velocity on this graph? The average velocity can be represented as the slope of the secant line through the initial and final points.   from time 0 to time 3.5 s(t) = Height off of the ground(in feet) t = Time in seconds …can be shown graphically to be…

Find Mr. Murphy’s average velocity between 1 and 3 seconds. =  64 feet/sec …the slope of the line through the initial and final points.   from time 1 to time 3 s(t) = Height off of the ground(in feet) t = Time in seconds

Approximate Mr. Murphy’s instantaneous (exact) velocity at 3 seconds. =  80 feet/sec …which is close to the exact velocity at 3 seconds.   from time 2 to time 3 We can draw a secant line close to 3. we’ll start with 2 seconds. s(t) = Height off of the ground(in feet) t = Time in seconds

Approximate Mr. Murphy’s instantaneous (exact) velocity at 3 seconds. =  88 feet/sec …which is even closer to the exact velocity at 3 seconds.   from time 2.5 to time 3 We can even try a secant line through 2.5 and 3. s(t) = Height off of the ground(in feet) t = Time in seconds

I think we’re getting the idea of how to find Mr. Murphy’s instantaneous (exact) velocity at 3 seconds.  We would need to get the secant points as close as we can. How would we do that? By taking the LIMIT as one point approaches the other…  s(t) = Height off of the ground(in feet) t = Time in seconds

 By taking the LIMIT as one point approaches the other… In other words… s(t) = Height off of the ground(in feet) t = Time in seconds We would need to get the secant points as close as we can. How would we do that? I think we’re getting the idea of how to find Mr. Murphy’s instantaneous (exact) velocity at 3 seconds.

So from this we find that as the two points on a secant line approach each other, it becomes the tangent line. And the slope of the tangent line is also called… The Derivative Let …which can also be written as… And since Hey! This is from Algebra class!

   f(x) x