Functions - AP Calculus AB
Card 0 of 43260
You wish to find the area under the curve 
between the values of 
and 
. Which is the correct way to express this task as an equation?
You wish to find the area under the curve
To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 2. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 6. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.
To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 2. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 6. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.
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You wish to find the area under the line 
between the values of 
and 
. Which is the correct way to express this task as an equation?
You wish to find the area under the line
To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 3. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 100. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.
To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 3. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 100. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.
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You wish to find the area under the line 
between the values of 
and 
. Which is the correct way to express this task as an equation?
You wish to find the area under the line
To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 4. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 12. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.
To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 4. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 12. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.
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You wish to find the area under the curve 
between the values of 
and 
. Which is the correct way to express this task as an equation?
You wish to find the area under the curve
To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 5. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 8. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.
To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 5. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 8. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.
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Find f'(1) given:
Find f'(1) given:
To solve, you must first differentiate according to the power rule and then plug in 1. Thus,
Power rule:
Since for all values of x, the first derivative is 6, our answer is 6.
To solve, you must first differentiate according to the power rule and then plug in 1. Thus,
Power rule:
Since for all values of x, the first derivative is 6, our answer is 6.
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Determine the slope of the line that is tangent to the function 
at the point 
Determine the slope of the line that is tangent to the function
The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.
We'll need to make use of the following derivative rule(s):
Trigonometric derivative:
Note that u may represent large functions, and not just individual variables!
Taking the derivative of the function
The slope of the tangent is
The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.
We'll need to make use of the following derivative rule(s):
Trigonometric derivative:
Note that u may represent large functions, and not just individual variables!
Taking the derivative of the function
The slope of the tangent is
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There are equations in physics that can tell you about the position and velocity of an object as a function of time. If we were to graph the equation of the relationship between the object's height and time it would be a quadratic while graphing the velocity vs time would be linear (since velocity is the derivative of position in respect to time. The equation of the height of the projectile with respect to time is the following
where 
is the initial height, 
is the acceleration, and 
is the initial velocity.
Say you throw a ball straight up in the air at an initial velocity of 20m/s. At approximately what time does the ball reach it's highest height? Use 
as the approximation for the acceleration due to gravity.
There are equations in physics that can tell you about the position and velocity of an object as a function of time. If we were to graph the equation of the relationship between the object's height and time it would be a quadratic while graphing the velocity vs time would be linear (since velocity is the derivative of position in respect to time. The equation of the height of the projectile with respect to time is the following
Say you throw a ball straight up in the air at an initial velocity of 20m/s. At approximately what time does the ball reach it's highest height? Use
Given that the relationship between height and time is a parabola, we note that the greatest height will occur when the first derivative is zero.
Given the equation,
The derivative is
which we then set equal to zero and plug in our known values.
We get the following:
.
Then solving for time, t, gives us 2.04s.
Given that the relationship between height and time is a parabola, we note that the greatest height will occur when the first derivative is zero.
Given the equation,
The derivative is
which we then set equal to zero and plug in our known values.
We get the following:
Then solving for time, t, gives us 2.04s.
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As per the mean value theorem, there exists at least one value 
within the interval 
such that 
. For the function, interval, and derivative value 
, 
, 
, find a value of 
that validates the mean value theorem.
As per the mean value theorem, there exists at least one value
The mean value theorem states that for a planar arc passing through a starting and endpoint 
, there exists at a minimum one point, 
, within the interval 
for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.
In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.
Using our function, interval, and derivative definitions, 
, 
, 
, we'll in turn wish to solve for 
to validate the mean value theorem:
Solving for 
using a calculator gives the solution:
There are two solutions to the equation; elect the value larger than 
to allow creation of an interval:
The mean value theorem states that for a planar arc passing through a starting and endpoint
In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.
Using our function, interval, and derivative definitions,
Solving for
There are two solutions to the equation; elect the value larger than
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As per the mean value theorem, there exists at least one value within the interval 
such that 
. For the function, interval, and derivative value 
, 
, 
, find a value of 
that validates the mean value theorem.
As per the mean value theorem, there exists at least one value
The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.
In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.
Using our function, interval, and derivative definitions, 
, 
, 
, we'll in turn wish to solve for 
to validate the mean value theorem:
Solving for 
using a calculator gives the solution:
To enable the creation of an interval, elect the value greater than a:
The mean value theorem states that for a planar arc passing through a starting and endpoint
In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.
Using our function, interval, and derivative definitions,
Solving for
To enable the creation of an interval, elect the value greater than a:
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As per the mean value theorem, there exists at least one value within the interval 
such that 
. For the function, interval, and derivative value 
, 
, 
, find a value of 
that validates the mean value theorem.
As per the mean value theorem, there exists at least one value
The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.
In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.
Using our function, interval, and derivative definitions, 
, 
, 
, we'll in turn wish to solve for 
to validate the mean value theorem:
Solving for 
using a calculator gives the solution:
Elect the value that is greater than a to complete the interval:
The mean value theorem states that for a planar arc passing through a starting and endpoint
In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.
Using our function, interval, and derivative definitions,
Solving for
Elect the value that is greater than a to complete the interval:
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As per the mean value theorem, there exists at least one value within the interval 
such that 
. For the function, interval, and derivative value 
, 
, 
, find a value of 
that validates the mean value theorem.
As per the mean value theorem, there exists at least one value
The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.
In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.
Using our function, interval, and derivative definitions, 
, 
, 
, we'll in turn wish to solve for 
to validate the mean value theorem:
Solving for 
using a calculator gives the solution:
Which is, as expected, greater than 
to allow the creation of an interval.
The mean value theorem states that for a planar arc passing through a starting and endpoint
In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.
Using our function, interval, and derivative definitions,
Solving for
Which is, as expected, greater than
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Consider a line tangent to the function 
at point 
. If this line also passes through point 
, then the following must be true:
Consider a line tangent to the function
The slope of the tangent line, 
, is equal to the slope of the function where it is defined. Furtheremore, the derivative of the function at this point is equal to the value of this slope:
We can find the slope of the tangent line in this problem using rise over run:
Therefore:
The slope of the tangent line,
We can find the slope of the tangent line in this problem using rise over run:
Therefore:
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Utilize the method of midpoint Riemann sums to approximate the average of 
over the interval 
using three midpoints.
Utilize the method of midpoint Riemann sums to approximate the average of
To find the average of a function over a given interval of values 
, the most precise method is to use an integral as follows:
Now for functions that are difficult or impossible to integrate,a Riemann sum can be used to approximate the value. A Riemann sum integral approximation over an interval with 
subintervals follows the form:
It is essentially a sum of 
rectangles each with a base of length equal to the subinterval length 
, and variable heights 
, which depend on the function value at a given point 
.
Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:
We're asked to approximate the average of 
over the interval 
The subintervals have length 
, and since we are using the midpoints of each interval, the x-values are 
To find the average of a function over a given interval of values
Now for functions that are difficult or impossible to integrate,a Riemann sum can be used to approximate the value. A Riemann sum integral approximation over an interval
It is essentially a sum of
Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:
We're asked to approximate the average of
The subintervals have length
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Consider a line tangent to the function 
at point 
. If this line also passes through point 
, then the following must be true:
Consider a line tangent to the function
The slope of the tangent line, 
, is equal to the slope of the function where it is defined. Furtheremore, the derivative of the function at this point is equal to the value of this slope:
We can find the slope of the tangent line in this problem using rise over run:
Therefore:
The slope of the tangent line,
We can find the slope of the tangent line in this problem using rise over run:
Therefore:
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Consider a line tangent to the function 
at point 
. If this line also passes through point 
, then the following must be true:
Consider a line tangent to the function
The slope of the tangent line, 
, is equal to the slope of the function where it is defined. Furtheremore, the derivative of the function at this point is equal to the value of this slope:
We can find the slope of the tangent line in this problem using rise over run:
Therefore:
The slope of the tangent line,
We can find the slope of the tangent line in this problem using rise over run:
Therefore:
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Find the critical points of
Find the critical points of
First we need to find 
.
Now we set 
Now we can use the quadratic equation in order to find the critical points.
Remember that the quadratic equation is
,
where a,b,c refer to the coefficients in the equation
In this case, a=3, b=6, and c=1.
Thus are critical points are
First we need to find
Now we set
Now we can use the quadratic equation in order to find the critical points.
Remember that the quadratic equation is
where a,b,c refer to the coefficients in the equation
In this case, a=3, b=6, and c=1.
Thus are critical points are
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Find the critical points of
.
Find the critical points of
In order to find the critical points, we need to find 
using the power rule 
.
Now we set 
, and solve for 
.
Thus 
is a critical point.
In order to find the critical points, we need to find
Now we set
Thus
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Find the critical point(s) of 
.
Find the critical point(s) of
To find the critical point(s) of a function 
, take its derivative 
, set it equal to 
, and solve for 
.
Given 
, use the power rule
to find the derivative. Thus the derivative is, 
.
Since 
:
The critical point 
is 
.
To find the critical point(s) of a function
Given
Since
The critical point
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A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its sides when its sides have length 24?
A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its sides when its sides have length 24?
Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:
The rates of change of the area can be found by taking the derivative of each side of the equation with respect to time:
Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the surface area and diagonal:
Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:
The rates of change of the area can be found by taking the derivative of each side of the equation with respect to time:
Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the surface area and diagonal:
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A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its sides when its sides have length 30?
A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its sides when its sides have length 30?
Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:
The rates of change of the area can be found by taking the derivative of each side of the equation with respect to time:
Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the surface area and diagonal:
Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:
The rates of change of the area can be found by taking the derivative of each side of the equation with respect to time:
Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the surface area and diagonal:
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