Calculus 2 : Graphing Parametrics

Study concepts, example questions & explanations for Calculus 2

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Example Questions

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Example Question #1 : Graphing Parametrics

Suppose we have a curve parameterized by the equations:

What is the tangent line to the curve at ?

Possible Answers:

Correct answer:

Explanation:

At , the graph passes through 

Now to find the slope, we will need both derivatives with respect to t, which are:

So to obtain the slope, we just use:

 ,

and evaluate at .

As it turns out,  at , and , so the slope will be 0 for this curve at the point .

That means that , and so solving at ordered pair , the solution must be:

Example Question #2 : Graphing Parametrics

Describe the graph of the following set of parametric equations:

Possible Answers:

An ellipse, centered at  with horizontal axis  and vertical axis .

An ellipse, centered at  with horizontal axis  and vertical axis .

A circle, centered at  with a radius of .

A circle, centered at  with a radius of .

A sinusoidal graph with amplitude , shifted up one unit and left two units.

Correct answer:

An ellipse, centered at  with horizontal axis  and vertical axis .

Explanation:

Perform these operations:

Now, we can use a Pythagorean trigonometric identity to transform the equation into a rectangular equation:

 

And this is the equation of an ellipse, centered at  with horizontal axis  and vertical axis .

 

 

Example Question #3 : Graphing Parametrics

Given  and , what is  in terms of  (rectangular form)?

Possible Answers:

None of the above

Correct answer:

Explanation:

Given  and  let's solve both equations for :

Since both equations equal , let's set them equal to each other and solve for :

 

Example Question #4 : Graphing Parametrics

Given  and , what is the length of the arc from ?

Possible Answers:

Correct answer:

Explanation:

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:

 

Example Question #5 : Graphing Parametrics

Find  using the following parametric equations

.

Possible Answers:

Correct answer:

Explanation:

It is known that we can derive  with the formula

So we just find :

To find these derivatives we will need to use the power rule, chain rule, and rule of exponentials.

Power Rule: 

Chain Rule: 

Rules of Exponentials: 

Applying these rules we get the following.

so we have 

.

Example Question #6 : Graphing Parametrics

Given the parametric equations

find .

Possible Answers:

Correct answer:

Explanation:

It is known that we can derive  with the formula

So we just find :

To find the derivatives we will need to use trigonometric rules and the rule for natural logs.

Trigonometric Rule for cosine: 

Rule of Natural Log: 

Applying the above rules we get the following derivatives.

so we have 

.

Example Question #7 : Graphing Parametrics

Graph the following parametric equation:

Possible Answers:

Y 1overx 2

Xsquaredplusysquared 1

Ysquaredminusxsquared 1

None of the other answers

Xsquaredminusysquared 1

Correct answer:

Xsquaredminusysquared 1

Explanation:

Using the identity , we can plug in the values for and  for to obtain the equation . This is the graph of a horizontal hyperbola with x-intercepts of and with asympotes of . The picture is depicted below:

Xsquaredminusysquared 1

Example Question #8 : Graphing Parametrics

In which quadrant does the parametric equation terminate when  ?

Possible Answers:

Correct answer:

Explanation:

When 

we have that

This gives the coordinate

which is located in

Example Question #9 : Graphing Parametrics

In which quadrant does the parametric equation terminate when  ?

Possible Answers:

Correct answer:

Explanation:

When 

we have that

This gives the coordinate

which is located in

 
 

Example Question #10 : Graphing Parametrics

In which quadrant is the parametric function located for the given  value?

Possible Answers:

Correct answer:

Explanation:

We substitute the given  value into the parametric function.

The resulting coordinate is in

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