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Calculus 2 : Graphing Parametrics

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Example Question #1 : Graphing Parametrics

Suppose we have a curve parameterized by the equations:

$x = e^{-t}$

y = t^3 - 3t^2 + 1

What is the tangent line to the curve at ?

Possible Answers:

y = -3

y = x + e

y = e^2 x - 4

y = -e x + 1

$y = -3x -3 + 3e^{-2}$

Correct answer:

y = -3

Explanation:

At t=2 , the graph passes through $(e^{-2}, 2^3 - 3\cdot 2^2 + 1) = (e^{-2}, -3)$

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

$\frac{dx}{dt} = -e^{-t}$

$\frac{dy}{dt} = 3t^2 - 6t$

So to obtain the slope, we just use:

$\frac{\mathrm{dy} }{\mathrm{d} x} = \frac{dy/dt}{dx/dt}$ ,

and evaluate at t=2 .

As it turns out, $\frac{\mathrm{dy} }{\mathrm{d} t}$ at t=2 , and $\frac{\mathrm{dx} }{\mathrm{d} t} \neq 0$ , so the slope will be 0 for this curve at the point t=2 .

That means that y = 0x +b = b , and so solving at ordered pair $(e^{-2}, -3)$ , the solution must be:

y = -3

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Example Question #101 : Parametric

Describe the graph of the following set of parametric equations:

x = 4 sin(t) - 2

y = 9cos(t) + 1

Possible Answers:

An ellipse, centered at (-2, 1) with horizontal axis and vertical axis .

A circle, centered at (2,-1) with a radius of $\frac{9}{8}$ .

An ellipse, centered at (-2, 1) with horizontal axis and vertical axis .

A circle, centered at (2,-1) with a radius of $\frac{9}{4}$ .

A sinusoidal graph with amplitude $\frac{2}{3}$ , shifted up one unit and left two units.

Correct answer:

An ellipse, centered at (-2, 1) with horizontal axis and vertical axis .

Explanation:

x = 4 sin(t) - 2

y = 9cos(t) + 1

Perform these operations:

$x + 2 = 4sin (t) \Rightarrow \frac{x+2}{4} = sin(t)$

$y - 1 = 9 cos(t) \Rightarrow \frac{y-1}{9} = cos(t)$

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

$\frac{(x+2)^2}{4^2} + \frac{(y-1)^2}{9^2} = sin^2t + cos^2 t = 1$

And this is the equation of an ellipse, centered at (-2, 1) with horizontal axis and vertical axis .

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Example Question #3 : Graphing Parametrics

Given x=-t+5 and y=9t+4 , what is in terms of (rectangular form)?

Possible Answers:

y=49-9x

y=9x-49

None of the above

y=49+9x

y=9x+49

Correct answer:

y=49-9x

Explanation:

Given x=-t+5 and y=9t+4 let's solve both equations for :

$x=-t+5\rightarrow t=5-x$

$y=9t+4\rightarrow t=\frac{y-4}{9}$

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

$\frac{y-4}{9}=5-x$

y-4=9(5-x)

y=9(5-x)+4

y=45-9x+4

y=49-9x

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Example Question #4 : Graphing Parametrics

Given x=2t+7 and y=7t-2 , what is the length of the arc from $0\leq t\leq2$ ?

Possible Answers:

$3\sqrt{53}$

$\sqrt{53}$

$6\sqrt{53}$

$2\sqrt{53}$

$5\sqrt{53}$

Correct answer:

$2\sqrt{53}$

Explanation:

In order to find the arc length, we must use the arc length formula for parametric curves:

$L=\int_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}dt$ .

Given x=2t+7 and y=7t-2 , we can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ ,

to derive

$\frac{dx}{dt}=(1)2t^{1-1}+(0)7=2$ and $\frac{dy}{dt}=(1)7t^{1-1}-(0)2=7$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{0}^{2}\sqrt{(2)^{2}+(7)^{2}}dt$

$L=\int_{0}^{2}(\sqrt{4+49})dt$

$L=\int_{0}^{2}(\sqrt{53})dt$

Now, using the Power Rule for Integrals $\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ , we can determine that:

$L=\int_{0}^{2}(\sqrt{53})dt$

$L=[t\sqrt{53}]_{0}^{2}\textrm{}$

$L=2\sqrt{53}-0\sqrt{53}$

$L=2\sqrt{53}$

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Example Question #2 : Graphing Parametrics

Find $\frac{dy}{dx}$ using the following parametric equations

$x(t)=t^2+t^3, y(t)=e^{2t}$ .

Possible Answers:

$\frac{dy}{dx}=\frac{2e^{2t}}{3t^2+2t}$

$\frac{dy}{dx}=3t^2+2t$

$\frac{dy}{dx}=2e^{2t}$

$\frac{dy}{dx}=\frac{3t^2+2t}{2e^{2t}}$

Correct answer:

$\frac{dy}{dx}=\frac{2e^{2t}}{3t^2+2t}$

Explanation:

It is known that we can derive $\frac{dy}{dx}$ with the formula

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

So we just find $\frac{dy}{dt}, \frac{dx}{dt}$ :

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

Power Rule: $x^n\ dx =nx^{n-1}$

Chain Rule: $f(g(x))\ dx=f'(g(x))\cdot g'(x)$

Rules of Exponentials: $e^u dx=e^u \cdot \frac{du}{dx}$

Applying these rules we get the following.

$\frac{dy}{dt}=2e^{2t}$

$\frac{dx}{dt}=3t^2+2t$

so we have

$\frac{dy}{dx}=\frac{2e^{2t}}{3t^2+2t}$ .

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Example Question #101 : Parametric

Given the parametric equations

$x(t)=\cos t, y(t)=\ln t$

find $\frac{dy}{dx}$ .

Possible Answers:

$\frac{dy}{dx}=-\frac{t}{\sin t}$

$\frac{dy}{dx}=-\frac{1}{t\sin t}$

$\frac{dy}{dx}=-\sin t$

$\frac{dy}{dx}=-t\sin t$

Correct answer:

$\frac{dy}{dx}=-\frac{1}{t\sin t}$

Explanation:

It is known that we can derive $\frac{dy}{dx}$ with the formula

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

So we just find $\frac{dy}{dt}, \frac{dx}{dt}$ :

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

Trigonometric Rule for cosine: cos(t)dt=-sin(t)

Rule of Natural Log: $ln(u)dt=\frac{1}{u}\cdot \frac{du}{dt}$

Applying the above rules we get the following derivatives.

$\frac{dy}{dt}=\frac{1}{t}$

$\frac{dx}{dt}=-\sin t$

so we have

$\frac{dy}{dx}=-\frac{1}{t\sin t}$ .

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Example Question #101 : Parametric, Polar, And Vector

Graph the following parametric equation:

$x=\sec(t), y=\tan(t), t\in(-\infty,\infty)$

Possible Answers:

Xsquaredplusysquared 1

Xsquaredminusysquared 1

Ysquaredminusxsquared 1

None of the other answers

Y 1overx 2

Correct answer:

Xsquaredminusysquared 1

Explanation:

Using the identity $\sec^2(t)-\tan^2(t)=1$ , we can plug in the values for $\sec(t)$ and for $\tan(t)$ to obtain the equation x^2-y^2=1 . This is the graph of a horizontal hyperbola with x-intercepts of (-1,0) and (1,0) with asympotes of $y=\pm{x}$ . The picture is depicted below:

Xsquaredminusysquared 1

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Example Question #1 : Graphing Parametrics

In which quadrant does the parametric equation terminate when $t=\frac{\pi}{4}$ ?

$x=cos\,t$

$y=sin\,t$

Possible Answers:

III

Correct answer:

Explanation:

When $t=\frac{\pi}{4}$

we have that

$x=cos(\frac{\pi}{4})=\frac{\sqrt2}{2}$

$y=sin(\frac{\pi}{4})=\frac{\sqrt2}{2}$

This gives the coordinate

$(\frac{\sqrt2}{2},\frac{\sqrt2}{2})$

which is located in

$Quadrant\, I$

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Example Question #1 : Graphing Parametrics

In which quadrant does the parametric equation terminate when t=3 ?

x=8t

y=-4t

Possible Answers:

III

Correct answer:

Explanation:

When t=3

we have that

x=8(3)=24

y=-4(3)=-12

This gives the coordinate

(24,-12)

which is located in

$Quadrant\, IV$

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Example Question #10 : Graphing Parametrics

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

$(3t,\, t^3)$

t=-2

Possible Answers:

III

Correct answer:

III

Explanation:

We substitute the given value into the parametric function.

$(3(-2), \,(-2)^3)$

=(-6,-8)

The resulting coordinate is in

$Quadrant\,III$

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Calculus 2 : Graphing Parametrics

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Graphing Parametrics

Example Question #101 : Parametric

Example Question #3 : Graphing Parametrics

Example Question #4 : Graphing Parametrics

Example Question #2 : Graphing Parametrics

Example Question #101 : Parametric

Example Question #101 : Parametric, Polar, And Vector

Example Question #1 : Graphing Parametrics

Example Question #1 : Graphing Parametrics

Example Question #10 : Graphing Parametrics

All Calculus 2 Resources

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