Parametric, Polar, and Vector Functions - AP Calculus BC
Card 0 of 984
Rewrite as a Cartesian equation:
![x = t^{2} + 2t + 1, y = t^{2} - 2t + 1, t \in [-1, 1]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/180244/gif.latex)
Rewrite as a Cartesian equation:



So
or 
We are restricting
to values on
, so
is nonnegative; we choose
.
Also,



So
or 
We are restricting
to values on
, so
is nonpositive; we choose

or equivalently,

to make
nonpositive.
Then,

and





So
or
We are restricting to values on
, so
is nonnegative; we choose
.
Also,
So
or
We are restricting to values on
, so
is nonpositive; we choose
or equivalently,
to make nonpositive.
Then,
and
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Rewrite as a Cartesian equation:

Rewrite as a Cartesian equation:

, so

This makes the Cartesian equation
.
, so
This makes the Cartesian equation
.
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Draw the graph of
from
.
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
.
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
.
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Draw the graph of
where
.
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.
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.
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Graph
where
.
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
.
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
.
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Draw the curve of
from
.
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
.
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
.
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If
and
, what is
in terms of
(rectangular form)?
If and
, what is
in terms of
(rectangular form)?
Given
and
, we can find
in terms of
by isolating
in both equations:


Since both of these transformations equal
, we can set them equal to each other:




Given and
, we can find
in terms of
by isolating
in both equations:
Since both of these transformations equal , we can set them equal to each other:
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Find the length of the following parametric curve
,
,
.
Find the length of the following parametric curve
,
,
.
The length of a curve is found using the equation 
We use the product rule,
, when
and
are functions of
,
the trigonometric rule,
and ![\frac{d}{dt}[sin(t)]=cos(t)](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/429206/gif.latex)
and exponential rule,
to find
and
.
In this case
, 





The length of this curve is

Using the identity 



Using the identity 

Using the trigonometric identity
where
is a constant and 

Using the exponential rule, 

Using the exponential rule,
, gives us the final solution

The length of a curve is found using the equation
We use the product rule,
, when
and
are functions of
,
the trigonometric rule,
and
and exponential rule,
to find
and
.
In this case
,
The length of this curve is
Using the identity
Using the identity
Using the trigonometric identity where
is a constant and
Using the exponential rule,
Using the exponential rule, , gives us the final solution
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Given
and
, what is the arc length between
?
Given and
, what is the arc length between
?
In order to find the arc length, we must use the arc length formula for parametric curves:
.
Given
and
, we can use using the Power Rule
for all
, to derive
and
.
Plugging these values and our boundary values for
into the arc length equation, we get:



Now, using the Power Rule for Integrals
for all
,
we can determine that:
![L=[t\sqrt{5}]_{0}^{4}\textrm{}dt](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/417526/gif.latex)


In order to find the arc length, we must use the arc length formula for parametric curves:
.
Given and
, we can use using the Power Rule
for all
, to derive
and
.
Plugging these values and our boundary values for into the arc length equation, we get:
Now, using the Power Rule for Integrals
for all
,
we can determine that:
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Given
and
, what is the length of the arc from
?
Given and
, what is the length of the arc from
?
In order to find the arc length, we must use the arc length formula for parametric curves:
.
Given
and
, we can use using the Power Rule
for all
, to derive
and
.
Plugging these values and our boundary values for
into the arc length equation, we get:





Now, using the Power Rule for Integrals
for all
,
we can determine that:
![L=[2t\sqrt{13}]_{0}^{2}\textrm{}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/420925/gif.latex)


In order to find the arc length, we must use the arc length formula for parametric curves:
.
Given and
, we can use using the Power Rule
for all
, to derive
and
.
Plugging these values and our boundary values for into the arc length equation, we get:
Now, using the Power Rule for Integrals
for all
,
we can determine that:
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Find dy/dx at the point corresponding to the given value of the parameter without eliminating the parameter:

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.
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.
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Rewrite in polar form:

Rewrite in polar form:
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Graph the equation
where
.
Graph the equation where
.
At angle
the graph as a radius of
. As it approaches
, the radius approaches
.
As the graph approaches
, the radius approaches
.
Because this is a negative radius, the curve is drawn in the opposite quadrant between
and
.
Between
and
, the radius approaches
from
and redraws the curve in the first quadrant.
Between
and
, the graph redraws the curve in the fourth quadrant as the radius approaches
from
.
At angle the graph as a radius of
. As it approaches
, the radius approaches
.
As the graph approaches , the radius approaches
.
Because this is a negative radius, the curve is drawn in the opposite quadrant between and
.
Between and
, the radius approaches
from
and redraws the curve in the first quadrant.
Between and
, the graph redraws the curve in the fourth quadrant as the radius approaches
from
.
Compare your answer with the correct one above
Draw the graph of
from
.
Draw the graph of from
.
Because this function has a period of
, the x-intercepts of the graph
happen at a reference angle of
(angles halfway between the angles of the axes).
Between
and
the radius approaches
from
.
Between
and
, the radius approaches
from
and is drawn in the opposite quadrant, the third quadrant because it has a negative radius.
From
to
the radius approaches
from
, and is drawn in the fourth quadrant, the opposite quadrant.
Between
and
, the radius approaches
from
.
From
and
, the radius approaches
from
.
Between
and
, the radius approaches
from
. Because it is a negative radius, it is drawn in the opposite quadrant, the first quadrant.
Then between
and
the radius approaches
from
and is draw in the second quadrant.
Finally between
and
, the radius approaches
from
.
Because this function has a period of , the x-intercepts of the graph
happen at a reference angle of
(angles halfway between the angles of the axes).
Between and
the radius approaches
from
.
Between and
, the radius approaches
from
and is drawn in the opposite quadrant, the third quadrant because it has a negative radius.
From to
the radius approaches
from
, and is drawn in the fourth quadrant, the opposite quadrant.
Between and
, the radius approaches
from
.
From and
, the radius approaches
from
.
Between and
, the radius approaches
from
. Because it is a negative radius, it is drawn in the opposite quadrant, the first quadrant.
Then between and
the radius approaches
from
and is draw in the second quadrant.
Finally between and
, the radius approaches
from
.
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What is the following coordinate in polar form?

Provide the angle in degrees.
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.
.
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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What is the equation
in polar form?
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
,




We can convert from rectangular form to polar form by using the following identities: and
. Given
, then
.
. Dividing both sides by
,
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What is the equation
in polar form?
What is the equation in polar form?
We can convert from rectangular form to polar form by using the following identities:
and
. Given
, then
. Multiplying both sides by
,



We can convert from rectangular form to polar form by using the following identities: and
. Given
, then
. Multiplying both sides by
,
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Convert the following function into 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.



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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What is the equation
in polar form?
What is the equation in polar form?
We can convert from rectangular to polar form by using the following trigonometric identities:
and
. Given
, then:


Dividing both sides by
, we get:




We can convert from rectangular to polar form by using the following trigonometric identities: and
. Given
, then:
Dividing both sides by , we get:
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What is the polar form of
?
What is the polar form of ?
We can convert from rectangular to polar form by using the following trigonometric identities:
and
. Given
, then:


Dividing both sides by
, we get:




We can convert from rectangular to polar form by using the following trigonometric identities: and
. Given
, then:
Dividing both sides by , we get:
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