All AP Calculus BC Resources
Example Questions
Example Question #1 : Parametric Form
Find dy/dx at the point corresponding to the given value of the parameter without eliminating the parameter:
The formula for dy/dx for parametric equations is given as:
From the problem statement:
If we plug these into the above equation we end up with:
If we plug in our given value for t, we end up with:
This is one of the answer choices.
Example Question #7 : Parametric, Polar, And Vector Functions
Draw the graph of from .
Between and , the radius approaches from .
From to the radius goes from to .
Between and , the curve is redrawn in the opposite quadrant, the first quadrant as the radius approaches .
From and , the curve is redrawn in the second quadrant as the radius approaches from .
Example Question #8 : Parametric, Polar, And Vector Functions
Draw the graph of where .
Because this function has a period of , the amplitude of the graph appear at a reference angle of (angles halfway between the angles of the axes).
Between and the radius approaches 1 from 0.
Between and , the radius approaches 0 from 1.
From to the radius approaches -1 from 0 and is drawn in the opposite quadrant, the fourth quadrant because it has a negative radius.
Between and , the radius approaches 0 from -1, and is also drawn in the fourth quadrant.
From and , the radius approaches 1 from 0. Between and , the radius approaches 0 from 1.
Then between and the radius approaches -1 from 0. Because it is a negative radius, it is drawn in the opposite quadrant, the second quadrant. Likewise, as the radius approaches 0 from -1. Between and , the curve is drawn in the second quadrant.
Example Question #781 : Calculus Ii
Graph where .
Taking the graph of , we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.
This leaves us with the areas from to , to , and to .
Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of .
To draw the graph, the radius is 1 at and traces to 0 at . As well, the negative part of the radius starts at -1 and traces to zero in the opposite quadrant, the third quadrant.
From to , the curves are traced from 0 to 1 and 0 to -1 in the fourth quadrant. Following this pattern, the graph is redrawn again from the areas included in to .
Example Question #201 : Ap Calculus Bc
Draw the curve of from .
Taking the graph of , we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.
This leaves us with the areas from to and to .
Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of .
To draw the graph, the radius is 0 at and traces to 1 at . As well, the negative part of the radius starts at 0 and traces to-1 in the opposite quadrant, the third quadrant.
From to , the curves are traced from 1 to 0 and -1 to 0 in the third quadrant.
Following this pattern, the graph is redrawn again from the areas included in to .
Example Question #161 : Parametric, Polar, And Vector
Rewrite in polar form:
Example Question #1 : Polar Form
What is the following coordinate in polar form?
Provide the angle in degrees.
To calculate the polar coordinate, use
However, keep track of the angle here. 68 degree is the mathematical equivalent of the expression, but we know the point (-2,-5) is in the 3rd quadrant, so we have to add 180 to it to get 248.
Some calculators might already have provided you with the correct answer.
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Example Question #201 : Ap Calculus Bc
What is the equation in polar form?
We can convert from rectangular form to polar form by using the following identities: and . Given , then .
. Dividing both sides by ,
Example Question #3 : Polar Form
What is the equation in polar form?
None of the above
We can convert from rectangular form to polar form by using the following identities: and . Given , then . Multiplying both sides by ,
Example Question #4 : Polar Form
Convert the following function into polar form:
The following formulas were used to convert the function from polar to Cartestian coordinates:
Note that the last formula is a manipulation of a trignometric identity.
Simply replace these with x and y in the original function.
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