Simple rate of change problem?

[physphys]

Homework Statement

The Function S(t) = 6t(t+1) describes the distance (in kilometers) that a car has traveled after a time t ( in hours), t E [0,6]

Find the average velocity of the car from t= 2 to t= 2 + ∆t

Homework Equations

y'= f(x+h) - f(x)
------------
h

The Attempt at a Solution

I know from the power rule that the derivative of this function is 12t + 6
however when I try to use the rate of change formula I get 6t+ 20

Here is my work

s(t)' = s(2 + ∆t) - s(2)
----------------
∆t
=6(2+∆t)(∆t+2+1) - 36
---------------------
∆t
= (12+ 6∆t)(∆t+3) -36
--------------------
∆t
= 12∆t + 36 + 6∆t^2 + 18t - 36
------------------------------
∆t
=6∆t^2 + 30t
------------
∆t

= 6t + 30

So where did I go wrong?

 

[Dick]

Homework Statement

The Function S(t) = 6t(t+1) describes the distance (in kilometers) that a car has traveled after a time t ( in hours), t E [0,6]

Find the average velocity of the car from t= 2 to t= 2 + ∆t

Homework Equations

y'= f(x+h) - f(x)
------------
h

The Attempt at a Solution

I know from the power rule that the derivative of this function is 12t + 6
however when I try to use the rate of change formula I get 6t+ 20

Here is my work

s(t)' = s(2 + ∆t) - s(2)
----------------
∆t
=6(2+∆t)(∆t+2+1) - 36
---------------------
∆t
= (12+ 6∆t)(∆t+3) -36
--------------------
∆t
= 12∆t + 36 + 6∆t^2 + 18t - 36
------------------------------
∆t
=6∆t^2 + 30t
------------
∆t

= 6t + 30

So where did I go wrong?

You didn't do anything wrong except for some typos and you changed Δt into t. You should get 6Δt+30. To compare that with the derivative you need to let Δt->0. So the derivative S'(2) should come out to 30. If you put t=2 into the derivative you computed, 12t+6, you'll get the same number.

 

[Ray Vickson]

Homework Statement

The Function S(t) = 6t(t+1) describes the distance (in kilometers) that a car has traveled after a time t ( in hours), t E [0,6]

Find the average velocity of the car from t= 2 to t= 2 + ∆t

Homework Equations

y'= f(x+h) - f(x)
------------
h

The Attempt at a Solution

I know from the power rule that the derivative of this function is 12t + 6
however when I try to use the rate of change formula I get 6t+ 20

Here is my work

s(t)' = s(2 + ∆t) - s(2)
----------------
∆t
=6(2+∆t)(∆t+2+1) - 36
---------------------
∆t
= (12+ 6∆t)(∆t+3) -36
--------------------
∆t
= 12∆t + 36 + 6∆t^2 + 18t - 36
------------------------------
∆t
=6∆t^2 + 30t
------------
∆t

= 6t + 30

So where did I go wrong?

S(t) = 6*t^2 + 6*t, so the instantaneous velocity is v(t) = dS(t)/dt = 12*t + 6, and for small Δt > 0 that will also be the average velocity over [t,t+Δt]. For t = 2 this gives average_velocity = 24 + 6 = 30 (km/hr). Your error was in writing 12Δt + 18Δt as 12Δt + 18t.

 

[physphys]

oh okay I see, for some reason I thought using the first principle and finding Y' would give me the derivative, however, it just gives me the rate of change. Thanks for helping!

 

[Dick]

oh okay I see, for some reason I thought using the first principle and finding Y' would give me the derivative, however, it just gives me the rate of change. Thanks for helping!

Right. You can get the derivative from what you did by replacing 2 with t, working through the algebra again and then taking the limit as Δt->0.