AP Calculus AB : Derivatives
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Example Questions
Example Question #31 : Derivative Interpreted As An Instantaneous Rate Of Change
A car going up a hill can be described by the following equation:
How fast is the car moving at the time
The cars velocity is the instantaneous rate of change of the position with respect to time, 
Evaluated at 
Example Question #82 : Concept Of The Derivative
Given that h(t) represents the height of a frisbee as a function of time, find the frisbee's velocity when 
Given that h(t) represents the height of a frisbee as a function of time, find the frisbee's velocity when 
To find the velocity function from a height or position function, we simply need to find the first derivative of the function. Then, to find the velocity at a given value of t, simply plug the value of t into the velocity function.
Recall the power rule:
This states that we can find the derivative of any polynomial by taking each term, multiplying by its exponent, and then subtracting one from the exponent.
Doing so yields:
Notice that our 14 dropped out. Any constant terms will drop our while differentiating.
Change it to v(t) and we are almost there.
We need v(1.5)
So, we get -264.75. This might seem like a strange velocity for a frisbee, but we don't know the units, so it might make sense within the relevant system.
Example Question #83 : Concept Of The Derivative
Given that h(t) represents the height of a frisbee as a function of time, find the frisbee's acceleration when 
Given that h(t) represents the height of a frisbee as a function of time, find the frisbee's acceleration when 
To find the acceleration function, we need to recall that velocity is the first derivative of position, and that acceleration is the first derivative of velocity. In other words, we need to find our second derivative. Once we have that, we need to plug in -1 for t.
Recall the power rule:
This states that we can find the derivative of any polynomial by taking each term, multiplying by its exponent, and then subtracting one from the exponent.
Doing so yields:
Notice that our 14 dropped out. Any constant terms will drop our while differentiating.
Change it to v(t) and we are almost there.
Next, find the derivative of v(t), again with the power rule.
Making our acceleration function:
Now, finally, plug in -1 for t and solve.
So, our acceleration is:
-754
We have no units, so although this number may seem a little strange, we can;t really say much about it.
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