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Activity:
Fundamental Theorem of Calculus
Discuss and analyze Graphically
Position
Velocity (Proof for instantaneous vs. Avg)(FINISH FOR HW)
Acceleration
SLOPE
Intercept
Area
Objective:
You should be able to Define Describe Differentiate and Relate the above quantities from both a Physical and mathematical standpoint.
Key Skill: be able to apply the fundamental theorem of calculus to a novel situation
What is wrong with the proof below:
Mr Crane how are we getting pizza tomorrow?
ReplyDeleteHere are easy ways to evaluate limits in a derivative sense:
ReplyDeleteUsing the limit definition of a derivative we can find the instantaneous rate of change of a function (obviously), in order to know if you're correct, just match it to it's derivative.
For example, a function such as x(t)=3t would evaluate to 3. But how? We can use the definition of the derivative (dt=delta t not differential), lim dt->0 of (x(t+dt)-x(t))/dt.
x(t+dt)=3(t+dt)=3t+3dt
x(t)=3t
lim dt->0 of (3t+3dt-3t)/dt
lim dt->0 of 3dt/dt.
The limit goes to 3.
Does this make sense? Well we know that in order to get the instantaneous rate of change, we can just take the derivative, so dx/dt=3, so yes our answer does make sense.
Let's do another one, given that x(t)=10t^2 +6t
lim dt->0 of (x(t+dt)-x(t))/dt
x(t+dt)=10(t+dt)^2 +6(t+dt)
x(t+dt)=6t+10t^2 +6dt+20tdt+10dt^2
x(t)=10t^2 +6t
lim dt->0 of (x(t+dt)-x(t))/dt
lim dt->0 of (6t+10t^2 +6dt+20tdt+10dt^2 -10t^2-6t)/dt
lim dt->0 of (6dt+20tdt+10dt^2)/dt
lim dt->0 of (6dt+20tdt+10dt^2)/dt = dt(6+20t+10dt)/dt
dt(6+20t+10dt)/dt = 6+20t+10dt
lim dt->0 of (6+20t+10dt) = 6+20t
Does this answer make sense? Well the original function was 10t^2 +6t and of course the derivative is 20t+6, so yes our final answer does make sense.
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