1. MTH1150 Tangents and Their Slopes

2. Two Basic Goals For Calculus
First Goal
When given a function, we want to be able to determine the instantaneous rate of change at any point.
To accomplish this we use the concept of differentiation
Second Goal
Determine the area under a function.
To accomplish this we use the concept of integration – MTH1170

3. Definition Tangent Line:
A Tangent Line is the line through a pair of infinitely close pair of points on the curve.
The slope of this line is the instantaneous rate of change of the function.

4. Definition
Slope
A number that describes both the direction and the steepness of a line.
𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1
In order to calculate slope we need to have the ordered pair for two points found on the line.

5. The Tangent Problem
The Tangent Problem - We know that when defining a line two points are used to calculate slope and then to express the line using the point slope formula. But how can this be calculated when the points are infinitely close to one another?
The answer is limits!

6. Slope of a Tangent Line
When given a function, we want to be able to determine the instantaneous rate of change of this function at any point.
To accomplish this, first we need to express the slope of a secant line defined in the most general way possible.

7. Slope of a Tangent Line
We know that in order to calculate slope we need to have the ordered pair for two points found on the line. We begin by finding the general expression for this.
𝑥 1 =𝑥 𝑥 2 =𝑥+ℎ 𝑦 1 =𝑓(𝑥) 𝑦 2 =𝑓(𝑥+ℎ)
𝑃 1 𝑥 1 , 𝑦 1 = 𝑃 1 𝑥, 𝑓(𝑥)
𝑃 2 𝑥 2 , 𝑦 2 = 𝑃 2 𝑥+ℎ, 𝑓(𝑥+ℎ)

8. Slope of a Tangent Line Given:
𝑃 1 𝑥 1 , 𝑦 1 = 𝑃 1 𝑥, 𝑓(𝑥)
𝑃 2 𝑥 2 , 𝑦 2 = 𝑃 2 𝑥+ℎ, 𝑓(𝑥+ℎ)
Then the slope of the line connecting these two points will be:
𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 𝑓 𝑥+ℎ −𝑓(𝑥) 𝑥+ℎ−𝑥 = 𝑓 𝑥+ℎ −𝑓(𝑥) ℎ
Where h is the distance that separates the two points found on the function f(x).

9. Slope of a Tangent Line
Now we have the general equation for the slope of a secant line. How could we turn this into the slope of a tangent line?
We could set the distance separating these two points to zero.
𝑚= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 𝑓 𝑥+ℎ −𝑓(𝑥) 𝑥+ℎ−𝑥 = 𝑓 𝑥+0 −𝑓(𝑥) 0
To numerically compute the slope of this line we would need to divide by zero.

10. Slope of a Tangent Line
But we could take the limit as h approaches zero!
𝑚 𝑇𝑎𝑛 = lim ℎ→0 𝑓 𝑥+ℎ −𝑓(𝑥) ℎ
This function will return the slope of a tangent line for any value of x where f(x) is defined.

11. The Newton Quotient (Difference Equation)
This new function that we have obtained is called the Newton Quotient.
𝑓′(𝑥)= lim ℎ→0 𝑓 𝑥+ℎ −𝑓(𝑥) ℎ
Where:
𝑓′(𝑥) is the function obtained by solving the limit problem found above. This new function returns the slope of a tangent line found at a point on f(x).

12. Steps to Solving the NQ
The steps required to solve the Newton Quotient are as follows:
Steps:
Define f(x) and f(x+h)
Simplify f(x) and f(x+h)
Plug f(x) and f(x+h) into the NQ
Rearrange and simplify the numerator in order to factor out an h 
Use the h that you factored out in step 4 to divide out the h in the denominator
Solve the limit by plugging in h = 0

13. Examples
Find the new function f’(x) by running the following functions through the Newton Quotient:
𝑦=𝑥
𝑦= 𝑥 2
𝑦= 5𝑥 2
𝑦= 1 𝑥
𝑦= 𝑥