1.6 Slopes of Tangents & Instantaneous Rates of Change - Period 4

Hi Everyone,

My Name is David L I will be the scribe of today's post.


Today lesson is about Instantaneous Rates of Change:


What is Instantaneous Rates of Change? How do we find it?

Instantaneous Rate of Change is an approximation of how a graph changes at a accurate instance in time.

Alternative 1: Find the slope of secant

E.g. In this scenario Point X (10,10) and Point Y (20,22) are points that the secant passes through.

Slope = delta d - delta t
= D2-D1 / T2 - T1
= (22 - 10) / (20 - 10)
= 12 / 10
= 1.2

Alternative 2: Use 2 points on the secant line (one being the tangent point and other is a point passing through secant line)

E.g. In this situation Point A (60,70) is secant point and Point B (70,80) is and regular point passing the secant line)

Slope = delta d - delta t
= D2-D1 / T2 - T1
= (80 - 70) / (70 - 60)
= 10 / 10
= 1.2 m/s

Alternative 3: Use an equation and set intervals between each point e.g. [0.1, 0.01,0.001]

e.g. The Instantaneous rate of change at 5.5 seconds:

h(t)=-4.2t^2+10t+2
h(5.5)=-4.2(5.5)^2+10(5.5)+2
h(5.5)=-70.05m

h(t)=-4.2t^2+10t+2
h(5.6)=-4.2(5.6)^2+10(5.6)+2
h(5.6)=-73.712m

h(t)=-4.2t^2+10t+2
h(t)=-4.2t^2+10t+2
h(5.1)=-4.2(5.1)^2+10(5.1)+2
h(5.1)=--70.41

h(t)=4.2t^2+10t+2
h(t)=4.2t^2+10t+2
h(5.01)=4.2(5.01)^2+10(5.01)+2
h(5.01)=-70.09

AROC(5.5,5.6)
=(-73.71--70.05) / (5.6-5.5)
=36.6m/s

AROC(5.5,5.51)
=(-70.41--70.05) / (5.51-5.5)
=-36m/s

AROC(5.5,5.501)
=(-70.09--70.05) / (5.501-5.5)
=-40m/s

Therefore the slope of tangent or instantaneous rates of change would be about -40 m/s.