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Slope of Tangent Lines and Instantaneous Rates of Change, Study notes of Pre-Calculus

Sullivan UniversityPre-Calculus

The concept of the slope of a curve, specifically the slope of the line tangent to the curve and the instantaneous rate of change. It provides examples of how to find the slope of the line tangent to a function using the limit definition and shows how the slope of the curve is equal to the instantaneous rate of change at a given point.

Typology: Study notes

2021/2022

Uploaded on 09/12/2022

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Recall: Slope of Secant Line =Ave. Rate of Change
If we are looking at the graph of y=f(x), then
slope of secant line from x=ato x=b=rise
run
=โˆ†y
โˆ†x
=f(b)โˆ’f(a)
bโˆ’a
If we are looking at an object whose position is given by f(t), then
average r.o.c. from t=ato t=b=change in position
time
=f(tfinal)โˆ’f(tinitial)
tFinal โˆ’tinitial
=f(b)โˆ’f(a)
bโˆ’a.
Math 101-Calculus 1 (Sklensky) In-Class Work September 21, 2011 1 / 7
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Download Slope of Tangent Lines and Instantaneous Rates of Change and more Study notes Pre-Calculus in PDF only on Docsity!

Recall: Slope of Secant Line =Ave. Rate of Change

If we are looking at the graph of y = f (x), then slope of secant line from x = a to x = b = riserun = โˆ† โˆ†yx

= f^ (b b) โˆ’โˆ’^ fa^ (a)

If we are looking at an object whose position is given by f (t), then average r.o.c. from t = a to t = b = change in positiontime = f^ (t tfinal)^ โˆ’^ f^ (tinitial) Final โˆ’^ tinitial = f^ (b b)^ โˆ’โˆ’^ fa^ (a).

What Weโ€™re Really Interested In:

I (^) Slope of a Curve (i.e, slope of the line tangent to the curve)

I (^) Instantaneous Rate of Change

In Class Work

  1. Use the limit definition to find the slope of the line tangent to f (x) = x^2 โˆ’ 3 x at the following points. If you happen to have learned a short cut for this process in a past class, do NOT use it here. The point of this work is to practice with and understand the limit definition of the slope of the secant line. (a) x = 0 (b) x = โˆ’ 2 (c) An unspecified point x
  2. Find the equation of the line tangent to f (x) = x^2 โˆ’ 3 x at x = โˆ’2.

Solutions

  1. Use the limit definition to find the slope of the line tangent to f (x) = x^2 โˆ’ 3 x at: (a) x = 0

mtan = (^) hlimโ†’ 0 f^ (0 +^ h h)^ โˆ’^ f^ (0)

= (^) hlimโ†’ 0

[ (h)^2 โˆ’ 3(h)

] โˆ’

[ 02 โˆ’ 3(0)

]

h = (^) hlimโ†’ 0 h

(^2) โˆ’ 3 h h = (^) hlimโ†’ 0 h โˆ’ 3 = โˆ’ 3

Solutions

  1. Use the limit definition to find the slope of the line tangent to f (x) = x^2 โˆ’ 3 x at: (c) An unspecified point x

mtan = (^) hlimโ†’ 0 f^ (x^ +^ h h)^ โˆ’^ f^ (x)

= (^) hlimโ†’ 0

[ (x + h)^2 โˆ’ 3(x + h)

] โˆ’

[ x^2 โˆ’ 3(x)

]

h = (^) hlimโ†’ 0

(x (^2) + 2xh + h (^2) โˆ’ 3 x โˆ’ 3 h) (^) โˆ’ (x (^2) โˆ’ 3 x) h = (^) hlimโ†’ 02 xh^ +^ h

(^2) โˆ’ 3 h h = (^) hlimโ†’ 0 2 x + h โˆ’ 3 = 2 x โˆ’ 3

Solutions

  1. Find the equation of the line tangent to f (x) = x^2 โˆ’ 3 x at x = โˆ’2. Need: point and a slope Slope: mtan(โˆ’2) = โˆ’ 7 Point: Use point of tangency, (^ โˆ’ 2 , f (โˆ’2))^ = (โˆ’ 2 , 4 + 6) = (โˆ’ 2 , 10) Thus the tangent line is given by y โˆ’ 10 = โˆ’7(x + 2) =โ‡’ y = โˆ’ 7 x โˆ’ 4.