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Calculus 2 : Parametric, Polar, and Vector

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Example Questions

Example Question #131 : Parametric, Polar, And Vector

Find the derivative of the following parametric function at t=0 :

$x=t^2\cos(t)+3t$ ,

y=te^t

Possible Answers:

$\frac{1}{3}$

$-\frac{1}{3}$

$\infty$

Correct answer:

$\frac{1}{3}$

Explanation:

The derivative of a parametric function is given by the following:

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

So, we must find the derivative of each function at the t given:

y'=e^t+te^t , $x'=2t\cos(t)-t^2\sin(t)+3$

The derivatives were found using the following rules:

$\frac{d}{dx}(e^x)=e^x$ , $\frac{d}{dx}(x^n)=nx^{n-1}$ ,

$\frac{d}{dx}\sin(x)=\cos(x)$ ,

$\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

Next, plug in the given t=0 into each derivative function:

y'(0)=1 , x'(0)=3

Finally, divide $\frac{y'}{x'}$ to get a final answer of $\frac{1}{3}$ .

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Example Question #132 : Parametric, Polar, And Vector

Find $\frac{dy}{dx}$ for the following set of parametric equations for t = 2 .

$y = 5t^{2} + 3t$

$x = \ln t$

Possible Answers:

Does not Exist

$\frac{1}{46}$

$\frac{1}{23}$

Correct answer:

Explanation:

Finding $\frac{dy}{dx}$ of a parametric equation can be given by this formula:

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

So we must find $\frac{dy}{dt}$ and $\frac{dx}{dt}$ for when t = 2 .

$\frac{dy}{dt} = 10t + 3$ and $\frac{dx}{dt} = \frac{1}{t}$ and so

$\frac{dy}{dx} = 10t^{2} + 3t$ .

When you plug in t = 2 you get your answer .

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Example Question #133 : Parametric, Polar, And Vector

Find the derivative of the following parametric equation

$\mathbf{r}(t)=t^{3}\mathbf{i}+3ln(t)\mathbf{j}+5sin(t)\mathbf{k}$

Possible Answers:

Does not exist

$\mathbf{r'}(t)=3t^{2}\mathbf{i}+\frac{3}{t}\mathbf{j}+5cos(t)\mathbf{k}$

$\mathbf{r'}(t)=3t^{2}+\frac{3}{t}+5cos(t)$

$\mathbf{r'}(t)=2t^{2}\mathbf{i}+3t\mathbf{j}+5cos(t)\mathbf{k}$

$\mathbf{r'}(t)=t^{3}\mathbf{i}+3ln(t)\mathbf{j}+5sin(t)\mathbf{k}$

Correct answer:

$\mathbf{r'}(t)=3t^{2}\mathbf{i}+\frac{3}{t}\mathbf{j}+5cos(t)\mathbf{k}$

Explanation:

This parametric equation is described as the sum of three vectors. To find the derivative of a parametric equation, you must find the derivative of each vector, or if

$\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+x(t)\mathbf{k}$ then $\mathbf{r'}(t)=x'(t)\mathbf{i}+y'(t)\mathbf{j}+x'(t)\mathbf{k}$

The derivative of the first vector is found using the power rule,

$\frac{d}{dt}(t^{c})=ct^{c-1}$ where is a constant.

The derivative of the second vector is found using the natural logarithm rule,

$\frac{d}{dt}[ln(t)]=\frac{1}{t}$ .

The derivative of the third vector is found using one of the trigonmetric rules,

$\frac{d}{dt}[sin(t)]=cos(t)$ .

In this case:

$\mathbf{r'}(t)=\frac{d}{dt}(t^{3})\mathbf{i}+3\frac{d}{dt}[ln(t)]\mathbf{j}+5\frac{d}{dt}[sin(t)]\mathbf{k}$

$=3t^{2}\mathbf{i}+\frac{3}{t}\mathbf{j}+5cos(t)\mathbf{k}$

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Example Question #134 : Parametric, Polar, And Vector

Find the derivative of the following parametric equation

$\mathbf{r}(t)=t\mathbf{i}+3e^{t}\mathbf{j}+sin(\pi t)\mathbf{k}$

Possible Answers:

$\mathbf{r'}(t)=t\mathbf{i}+3e^{t}\mathbf{j}+sin(\pi t)\mathbf{k}$

$\mathbf{r'}(t)=\mathbf{i}+3e^{t}\mathbf{j}+\pi cos(\pi t)\mathbf{k}$

Does not exist

$\mathbf{r'}(t)=1+3e^{t}+\pi cos(\pi t)$

$\mathbf{r'}(t)=t\mathbf{i}+9e^{t}\mathbf{j}+\pi sin(\pi t)\mathbf{k}$

Correct answer:

$\mathbf{r'}(t)=\mathbf{i}+3e^{t}\mathbf{j}+\pi cos(\pi t)\mathbf{k}$

Explanation:

This parametric equation is described as the sum of three vectors. To find the derivative of a parametric equation, you must find the derivative of each vector, or if

$\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+x(t)\mathbf{k}$ , then $\mathbf{r'}(t)=x'(t)\mathbf{i}+y'(t)\mathbf{j}+x'(t)\mathbf{k}$

The derivative of the first vector is found using the power rule,

$\frac{d}{dt}(t)=1$ .

The derivative of the second vector is found using the exponential rule,

$\frac{d}{dt}(e^{t})=e^{t}$ .

The derivative of the third vector is found using one of the trigonmetric rules,

$\frac{d}{dt}[sin(at)]=acos(at)$ , where is a constant.

In this case:

$\mathbf{r'}(t)=\frac{d}{dt}(t)\mathbf{i}+3\frac{d}{dt}(e^{t})\mathbf{j}+\frac{d}{dt}[sin(\pi t)]\mathbf{k}$

$=\mathbf{i}+3e^{t}\mathbf{j}+\pi cos(\pi t)\mathbf{k}$

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Example Question #135 : Parametric, Polar, And Vector

Find the derivative of the following parametric equation

$\mathbf{r}(t)=2cos(5t)\mathbf{i}+5sin(7t)\mathbf{j}$

Possible Answers:

$\mathbf{r'}(t)=-10sin(5t)\mathbf{i}+35cos(7t)\mathbf{j}$

$\mathbf{r'}(t)=2cos(5t)\mathbf{i}+5sin(7t)\mathbf{j}$

$\mathbf{r'}(t)=-10cos(5t)\mathbf{i}+35sin(7t)\mathbf{j}$

Does not exist

$\mathbf{r'}(t)=-10sin(5t)+35cos(7t)$

Correct answer:

$\mathbf{r'}(t)=-10sin(5t)\mathbf{i}+35cos(7t)\mathbf{j}$

Explanation:

This parametric equation is described as the sum of three vectors. To find the derivative of a parametric equation, you must find the derivative of each vector, or if

$\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+x(t)\mathbf{k}$ , then $\mathbf{r'}(t)=x'(t)\mathbf{i}+y'(t)\mathbf{j}+x'(t)\mathbf{k}$

The derivative of the first and second vectors are found using the following trigonometric rules,

$\frac{d}{dt}[cos(at)]=-asin(at)$ and $\frac{d}{dt}[sin(bt)]=bcos(bt)$ ,

where and are constants.

In this case:

$\mathbf{r'}(t)=2\frac{d}{dt}[cos(5t)]\mathbf{i}+5\frac{d}{dt}[sin(7t)]\mathbf{j}$

$=[2*(-sin(5t))*5]\mathbf{i}+[5*cos(7t)*7]\mathbf{j}$

$=-10sin(5t)\mathbf{i}+35cos(7t)\mathbf{j}$

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Example Question #136 : Parametric, Polar, And Vector

Find $\frac{dy}{dx}$ when x=sin(7t) and y=cos(7t) .

Possible Answers:

$\frac{dy}{dx}=-7tan(7t)$

$\frac{dy}{dx}=-7sin(7t)\mathbf{i}+7cos(7t)\mathbf{j}$

$\frac{dy}{dx}=-7sin(7t)$

$\frac{dy}{dx}=-tan(7t)$

$\frac{dy}{dx}=7cos(7t)$

Correct answer:

$\frac{dy}{dx}=-tan(7t)$

Explanation:

If x=x(t) and y=y(t) , then we can use the chain rule to define $\frac{dy}{dx}$ as

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

We then use the following trigonometric rules,

$\frac{d}{dt}[cos(at)]=-asin(at)$ and $\frac{d}{dt}[sin(bt)]=bcos(bt)$ ,

where and are constants.

In this case:

$\frac{dy}{dt}=\frac{d}{dt}(cos(7t))=-7sin(7t)$ ,

and

$\frac{dx}{dt}=\frac{d}{dt}(sin(7t))=7cos(7t)$ ,

therefore

$\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=\frac{-7sin(7t)}{7cos(7t)}=-tan(7t)$ .

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Example Question #1 : Parametric Calculations

Calculate the length of the curve drawn out by the vector function ${\mathbf{v}(t)} = \left \langle t,\sqrt{2}}e^t,\frac{1}{2}e^{2t} \right \rangle,$ from 0<t<1 .

Possible Answers:

$\frac{1}{2}(e^2+1)$

ln(2) -1

e+1

None of the other answers.

e^2 +e^4

Correct answer:

$\frac{1}{2}(e^2+1)$

Explanation:

The formula for arc length of a parametric curve in space is $\int_{a}^{b} \sqrt{(\frac{d\mathbf{v}_1}{dt})^2 +(\frac{d\mathbf{v}_2}{dt})^2 +(\frac{d\mathbf{v}_3}{dt})^2} dt$ for a <t<b .

Taking derivatives of each of the vector function components and substituting the values into this formula gives

$\int_{0}^{1} \sqrt{1 + 2e^{2t} + e^{4t}} dt$

We need to recognize that underneath the square root we have a perfect square, and we can write it as

$\int_{0}^{1} \sqrt{(e^{2t} + 1)^2} dt$

Solving this we get

$\int_{0}^{1} e^{2t}+1 dt = [\frac{1}{2}e^{2t} +t]_0^1$ $= \bigg(\frac{1}{2}e^2+1\bigg)-\bigg(\frac{1}{2}\bigg) = \frac{1}{2}(e^2+1)$

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Example Question #2 : Parametric Calculations

Calculate $\frac{dy}{dx}$ at the point (1,2) on the curve defined by the parametric equations

x(t) = e^t , y(t)=ln(t+1)+t^2 +2

Possible Answers:

ln(2)

None of the other answers

ln(3)

Correct answer:

None of the other answers

Explanation:

The correct answer is .

We use the equation

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{1}{t+1}+2t+0}{e^t}$

But we need a value for to substitute into our derivative. We can obtain such a by setting x(t) =1, y(t) =2 as our given point suggests.

1 = e^t $\Rightarrow t =0$
$2 = ln(t+1)+t^2+2 \Rightarrow t=0$
Since our values of match, t=0 is our correcct value. Substituting this into the derivative and simplifying gives us our answer of

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Example Question #2 : Parametric Calculations

Which of the answers below is the equation obtained by eliminating the parametric from the following set of parametric equations?

x=t^2+3t+5

$y=\frac{3}{2}t-6$

Possible Answers:

$y=\frac{4}{9}x^2+\frac{22}{3}x+33$

$x=\frac{3}{2}y^2+16y+\frac{4}{3}$

$y=-\frac{14}{3}x^2+\frac{4}{7}x-4$

$y=\frac{3}{2}x^2+16x+\frac{4}{3}$

$x=\frac{4}{9}y^2+\frac{22}{3}y+33$

Correct answer:

$x=\frac{4}{9}y^2+\frac{22}{3}y+33$

Explanation:

When the problem asks us to eliminate the parametric, that means we want to somehow get rid of our variable t and be left with an equation that is only in terms of x and y. While the equation for x is a polynomial, making it more difficult to solve for t, we can see that the equation for y can easily be solved for t:

$y=\frac{3}{2}t-6$

$t=\frac{2}{3}y+4$

Now that we have an expression for t that is only in terms of y, we can plug this into our equation for x and simplify, and we will be left with an equation that is only in terms of x and y:

$x=t^2+3t+5=(\frac{2}{3}y+4)^2+3(\frac{2}{3}y+4)+5$

$x=\frac{4}{9}y^2+\frac{16}{3}y+16+2y+12+5$

$x=\frac{4}{9}y^2+\frac{22}{3}y+33$

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Example Question #3 : Parametric Calculations

Suppose x=cos(t) and y=sin(t) . Find the arc length from $0\leq t \leq \pi$ .

Possible Answers:

$\pi$

$\frac{\pi}{2}$

$\sqrt{\pi}$

$\sqrt{\frac{\pi}{2}}$

$2\pi$

Correct answer:

$\pi$

Explanation:

Write the arc length formula for parametric curves.

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

Find the derivatives. The bounds are given in the problem statement.

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

$\frac{dy}{dt}=cos(t)$

$L=\int_{0}^{\pi}\sqrt{(-sin(t))^2+(cos(t))^2}\:dt$

$L=\int_{0}^{\pi}\sqrt{sin(t)^2+cos(t)^2}\:dt$

$L=\int_{0}^{\pi}1\:dt =[t]_{0}^{\pi}= \pi$

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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #131 : Parametric, Polar, And Vector

Example Question #132 : Parametric, Polar, And Vector

Example Question #133 : Parametric, Polar, And Vector

Example Question #134 : Parametric, Polar, And Vector

Example Question #135 : Parametric, Polar, And Vector

Example Question #136 : Parametric, Polar, And Vector

Example Question #1 : Parametric Calculations

Example Question #2 : Parametric Calculations

Example Question #2 : Parametric Calculations

Example Question #3 : Parametric Calculations

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