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Calculus 2 : Derivatives of Parametrics

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Example Questions

Example Question #21 : Derivatives Of Parametrics

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

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

$x = \ln t$

Possible Answers:

$\frac{1}{46}$

$\frac{1}{23}$

Does not Exist

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 #22 : Derivatives Of Parametrics

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

Possible Answers:

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

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

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

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

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

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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Calculus 2 : Derivatives of Parametrics

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #21 : Derivatives Of Parametrics

Example Question #133 : Parametric, Polar, And Vector

Example Question #134 : Parametric, Polar, And Vector

Example Question #135 : Parametric, Polar, And Vector

Example Question #22 : Derivatives Of Parametrics

All Calculus 2 Resources

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