Tangent lines and velocity
Before derivatives exist, two concrete questions — steepness of a curve and speed of a moving object — force you to invent the limit.
The tangent problem
The slope of a curve at a point cannot be computed from two points the way a line's slope can, so we approximate. A secant line through the point and a nearby point has a computable slope, and as the nearby point slides closer, those secant slopes settle toward a single value. The tangent slope is defined as the limit of the secant slopes.
m_{\text{tan}} = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}Instantaneous velocity
Velocity asks the same question in motion language. Average velocity over a time interval is distance traveled divided by elapsed time, which is exactly a secant slope on a position graph. Shrinking the time interval to zero turns average velocity into instantaneous velocity at that instant.
v(a) = \lim_{h \to 0} \frac{s(a+h) - s(a)}{h}Worked example
Let s(t) = 3t^2 give position in meters after t seconds. The average velocity from t = 2 to t = 2 + h is (3(2+h)^2 - 12)/h, which simplifies to 12 + 3h. As h shrinks to 0, this approaches 12, so the instantaneous velocity at t = 2 is 12 meters per second.
v(2) = \lim_{h \to 0} (12 + 3h) = 12Flashcard
Slope of the secant line through two points
Check your understanding
A ball's height is s(t) = t^2 meters after t seconds. What is its instantaneous velocity at t = 3?