All AP Calculus AB Resources
Example Questions
Example Question #4 : Interpretation Of The Derivative As A Rate Of Change
If p(t) gives the position of a planet as a function of time, find the function which models the planet's velocity.
If p(t) gives the position of a planet as a function of time, find the function which models the planet's velocity.
Velocity is the first derivative of position.
Therefore, all we need to do to solve this problem is to find the first derivative.
We can do this via the power rule and the rule for differentiating sine.
1)
2)
So, we these rules in mind, we get:
So our final answer is:
Example Question #5 : Interpretation Of The Derivative As A Rate Of Change
If p(t) gives the position of a planet as a function of time, find the planet's velocity when t=0.
If p(t) gives the position of a planet as a function of time, find the planet's velocity when t=0.
Velocity is the first derivative of position. Therefore, all we need to do to solve this problem is to find the first derivative of p(t) and then plug in 0 for t and solve.
We can do this via the power rule and the rule for differentiating sine and cosine.
1)
2)
So, we these rules in mind, we get:
So our velocity function is:
Now, plug in 0 and simplify.
So, our answer is -11. We are not given any units, so we do not need to worry about them.
Example Question #4 : Interpretation Of The Derivative As A Rate Of Change
A factory producing pens wants to maximize its output; to do so, it needs to find the rate of change of pen production. Find this rate if pens are produced according to the following function:
The rate of change of a function is given by the derivative of that function. So, to find the rate of change of production, we must take the first derivative of the function for production, which is equal to
found using the following rules:
,
Example Question #6 : Interpretation Of The Derivative As A Rate Of Change
Concrete at a factory flows according to the following theoretical model:
What is the rate of change of the concrete flow?
The rate of change of the concrete flow is given by the first derivative of the concrete flow function:
and was found using the following rules:
, ,
Example Question #1 : Interpretation Of The Derivative As A Rate Of Change
Find the speed of the car at t=5 if its position is given by
To determine the speed of the car, we must take the first derivative of the position function, which gives us the rate of change of the position of the car - in other words, speed.
The derivative was found using the following rule:
,
Evaluating the derivative function at t=5, we get our speed as
Example Question #2 : Interpretation Of The Derivative As A Rate Of Change
A group of scientists use the following code for describing velocity and acceleration of particles:
, when both the velocity and acceleration are positive;
, when the velocity is positive, but the acceleration is negative;
, when the velocity is negative, but the acceleration is positive;
, when both the velocity and acceleration are negative.
What code - at t=2 - will the scientists use when describing a particle moving with a position function given by the following equation:
We need more information to determine the code used
To determine which code the scientists will use, we must find the velocity and acceleration of the particle, given by the first and second derivatives of the position function, respectively, evaluated at the given point.
The velocity and acceleration functions, therefore, are
The derivatives were found using the following rules:
,
Evaluated at t=2, we find that
When the velocity is positive, but the acceleration is negative, the code the scientists use is .
Example Question #3 : Interpretation Of The Derivative As A Rate Of Change
Find the velocity of the particle at x=0, its position given by the following function:
To determine the velocity of the particle, we must take the first derivative of the position function - in other words, the rate of change of the position is the velocity:
The derivative was found using the following rule:
To find the velocity at the given point, we simply plug in the value into the velocity function:
Example Question #31 : Applications Of Derivatives
The velocity of a particle in feet per second can be modeled by the function
What is the acceleration of this particle at 6 seconds?
The acceleration of a particle is its change in velocity; therefore, to find its acceleration at a particular point in time, it is necessary to take the derivative of the velocity function and evaluate it for that value. In short, we want .
Differentiate :
Apply the sum rule:
Apply the chain rule by setting :
This is the acceleration function; substitute 6 for :
Example Question #31 : Applications Of Derivatives
The velocity of a particle in feet after seconds can be modeled by the equation
.
At seconds, give the acceleration of this particle in feet per second squared.
The acceleration function of a particle with respect to time is the derivative of the velocity function of the particle.
Find this derivative by first using the product rule:
Apply the product rule:
Evaluate :
feet per second squared.
Example Question #31 : Applications Of Derivatives
Determine the velocity of a particle at x=10 when the position function is given by
To determine the velocity function - the rate of change of position - we must take the derivative of the position function:
The derivative was found using the following rules:
,
To find the velocity at a certain value of x - the instantaneous rate of change of position - we simply plug in the given value of x into the velocity function:
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