The Tangent Line Problem Part of section 1.1

Calculus centers around 2 fundamental problems: 1)The tangent line -- differential calculus 2) The area problem -- integral calculus

Tangent Line Problem One of the classic problems in calculus is the tangent line problem. You are probably very good at finding the slope of a line. Since the slope of a line (and line always implies straight in the world of math) is the same everywhere on the line, you could pick two points on the line and calculate the slope. With a curve, however, the slope is different depending where you are on the curve. You aren't able to just pick any old two points and calculate the slope. Think about the parabola, Whose graph looks like Pretend you are a little bug and this graph is a mountain. The slope of the mountain is different at each point on the mountain. Watch how easy/hard it looks for the bug to climb. Just look at the bug’s angle.

Without calculus, we could estimate the slope at a particular point by choosing an additional point close to our point in question and then drawing a line between the points and finding the slope. So if I was interested in finding out the slope at point P, I could estimate the slope by using a point Q, drawing a secant line (which crosses the graph in two places), and then finding the slope of that secant line. This is continued on the next slide. P

So, we get an approximate slope and sometimes just an estimate is fine. To get an even better approximation, we can move Q closer to P and that secant line begins to look more and more like a tangent line to the graph. I’ll draw a tangent line at P in orange. The tangent line and the graph share the same slope at point P. P Q

P Q If we could get the secant line to look more like the tangent line, the slope of the secant line would be more like the slope of the tangent line. You might be asking yourself, why is this helpful? Well, if you could zoom in on point P, you’d see that the tangent line and the graph look so similar, you can hardly tell them apart. So, by finding an approximate slope of the tangent line, you’d be finding an approximate slope of the graph at the point P. The approximate slope of the secant line at point P is defined as the rate of change at point P, or the ratio of change in y to the change in x near the point P.

(2, f(2)), (2.05, f(2.05))2.05 – 2 = 0.05f(2.05) – f(2) (2, f(2)), (2.04, f(2.04))2.04 – 2 = 0.04f(2.04) – f(2) (2, f(2)), (2.03, f(2.03))2.03 – 2 = 0.03f(2.03) – f(2) (2, f(2)), (2.02, f(2.02))2.02 – 2 = 0.02f(2.02) – f(2) (2, f(2)), (2.01, f(2.01))2.01 – 2 = 0.01f(2.01) – f(2) Problem Graph y = f (x) = x 2 – 1 How to interpret “the change in y ” and “the change in x ”? For example, the rate of change at some point, say x = 2 is considered as average rate of change at its neighbor. Change in x Change in y

In general, the rate of change at a single point x = c is considered as an average rate of change at its neighbor P(c, f(c)) and Q(c + h, f(c + h)) under a procedure of a secant line when its neighbor is approaching to but not equal to that single point, or, in some other format, it can be interpreted as (c, f(c))and (c + Δ x, f(c + Δ x )) or, the ratio of change in function value y f(c + Δ x ) – f(c) to the change in variable x, or, c + Δ x – c = Δ x

P Q Slope of secant line is the “average rate of change”

Instantaneous rate of change (Slope at a point) when  0 the will be approaching to a certain value. This value is the limit of the slope of the secant line and is called the rate of change at a single point AKA instantaneous rate of change.

2) The area problem- integral calculus Uses rectangles to approximate the area under a curve. Question: What is the area under a curve bounded by an interval? Problem Graph on

Similar to the way we deal with the “Rate of Change”, we partition the interval with certain amount of subinterval with or without equal length. Then we calculate the areas of these individual rectangles and sum them all together. That is the approximate area for the area under the curve bounded by the given interval. If we allow the process of partition of the interval goes to infinite, the ultimate result is the area under a curve bounded by an interval. Uses rectangles to approximate the area under a curve. Left Height Right Height

1)Use 4 subdivisions and draw the LEFT HEIGHTS 2) Use 4 subdivisions and draw the RIGHT HEIGHTS Problem Find the area of the graph on

With calculus, we are able to find the actual slope of the tangent line. Without it, we can only estimate.