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Calculus 2 : Polar

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #822 : Calculus Ii

What is the derivative of $r=7\cos\theta$ ?

Possible Answers:

$\frac{1+2\cos^{2}\theta}{2\sin\theta\cos\theta}$

$\frac{2-\cos^{2}\theta}{2\sin\theta\cos\theta}$

$\frac{1-2\cos^{2}\theta}{\sin\theta+2\cos\theta}$

$\frac{1-2\cos^{2}\theta}{2\sin\theta\cos\theta}$

$-\frac{1-2\cos^{2}\theta}{2\sin\theta\cos\theta}$

Correct answer:

$\frac{1-2\cos^{2}\theta}{2\sin\theta\cos\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=7\cos\theta$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=-7\sin\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}$

$\frac{dy}{dx}=\frac{-7\sin\theta\sin\theta+7\cos\theta\cos\theta}{-7\sin\theta\cos\theta-7\cos\theta\sin\theta}$

$\frac{dy}{dx}=\frac{-7(\sin^{2}\theta-\cos^{2}\theta)}{-14\sin\theta\cos\theta}$

$\frac{dy}{dx}=\frac{(\sin^{2}\theta-\cos^{2}\theta)}{2\sin\theta\cos\theta}$

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can deduce that $\cos^{2}\theta=1-\sin^{2}\theta$ . Swapping this into the numerator, we get:

$\frac{dy}{dx}=\frac{(1-\cos^{2}\theta-\cos^{2}\theta)}{2\sin\theta\cos\theta}$

$\frac{dy}{dx}=\frac{1-2\cos^{2}\theta}{2\sin\theta\cos\theta}$

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Example Question #315 : Parametric, Polar, And Vector

What is the derivative of $r=6\sin\theta-10$ ?

Possible Answers:

$\frac{12\cos\theta\sin\theta-10\cos\theta}{6+12\sin^{2}\theta+10\sin\theta}$

$\frac{12\cos\theta\sin\theta+10\cos\theta}{6-12\sin^{2}\theta+10\sin\theta}$

$\frac{12\cos\theta\sin\theta-10\cos\theta}{6-12\sin^{2}\theta+10\sin\theta}$

$\frac{12\cos\theta\sin\theta-10\cos\theta}{6-12\sin^{2}\theta-10\sin\theta}$

$\frac{12\cos\theta\sin\theta+10\cos\theta}{6+12\sin^{2}\theta+10\sin\theta}$

Correct answer:

$\frac{12\cos\theta\sin\theta-10\cos\theta}{6-12\sin^{2}\theta+10\sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=6\sin\theta-10$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=6\cos\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}$

$\frac{dy}{dx}=\frac{6\cos\theta\sin\theta+(6\sin\theta-10)\cos\theta}{6\cos\theta\cos\theta-(6\sin\theta-10)\sin\theta}$

$\frac{dy}{dx}=\frac{6\cos\theta\sin\theta+6\sin\theta\cos\theta-10\cos\theta}{6\cos\theta\cos\theta-6\sin\theta\sin\theta+10\sin\theta}$

$\frac{dy}{dx}=\frac{12\cos\theta\sin\theta-10\cos\theta}{6(\cos^{2}\theta-\sin^{2}\theta)+10\sin\theta}$

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can deduce that $\cos^{2}\theta=1-\sin^{2}\theta$ . Swapping this into the denominator, we get:

$\frac{dy}{dx}=\frac{12\cos\theta\sin\theta-10\cos\theta}{6(1-\sin^{2}\theta-\sin^{2}\theta)+10\sin\theta}$

$\frac{dy}{dx}=\frac{12\cos\theta\sin\theta-10\cos\theta}{6(1-2\sin^{2}\theta)+10\sin\theta}$

$\frac{dy}{dx}=\frac{12\cos\theta\sin\theta-10\cos\theta}{6-12\sin^{2}\theta+10\sin\theta}$

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Example Question #311 : Parametric, Polar, And Vector

What is the derivative of $r=2+5\sin\theta$ ?

Possible Answers:

$\frac{10\cos\theta\sin\theta-2\cos\theta}{5+10\sin^{2}\theta+2\sin\theta}$

$\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

$\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta+2\sin\theta}$

$\frac{10\cos\theta\sin\theta+2\cos\theta}{5+10\sin^{2}\theta-2\sin\theta}$

$\frac{10\cos\theta\sin\theta-2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

Correct answer:

$\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=2+5\sin\theta$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=5\cos\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}$

$\frac{dy}{dx}=\frac{(5\cos\theta)\sin\theta+(2+5\sin\theta)\cos\theta}{(5\cos\theta)\cos\theta-(2+5\sin\theta)\sin\theta}$

$\frac{dy}{dx}=\frac{5\cos\theta\sin\theta+2\cos\theta+5\sin\theta\cos\theta}{5\cos^{2}\theta-2\sin\theta-5\sin^{2}\theta}$

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5(\cos^{2}\theta-\sin^{2}\theta)-2\sin\theta}$

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can deduce that $\cos^{2}\theta=1-\sin^{2}\theta$ . Swapping this into the denominator, we get:

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5(1-\sin^{2}\theta-\sin^{2}\theta)-2\sin\theta}$

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5(1-2\sin^{2}\theta)-2\sin\theta}$

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

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Example Question #21 : Derivatives Of Polar Form

What is the derivative of $r=\sin\theta$ ?

Possible Answers:

$\frac{2\cos\theta\sin\theta}{1-2\sin^{2}\theta}$

$-\frac{2\cos\theta\sin\theta}{1-2\sin^{2}\theta}$

$\frac{2\cos\theta\sin\theta}{2-\sin^{2}\theta}$

$\frac{2\cos\theta\sin\theta}{1+2\sin^{2}\theta}$

$-\frac{2\cos\theta\sin\theta}{1+2\sin^{2}\theta}$

Correct answer:

$\frac{2\cos\theta\sin\theta}{1-2\sin^{2}\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=\sin\theta$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=\cos\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}$

$\frac{dy}{dx}=\frac{(\cos\theta)\sin\theta+(\sin\theta)\cos\theta}{(\cos\theta)\cos\theta-(\sin\theta)\sin\theta}$

$\frac{dy}{dx}=\frac{2\cos\theta\sin\theta}{\cos^{2}\theta-\sin^{2}\theta}$

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can deduce that $\cos^{2}\theta=1-\sin^{2}\theta$ . Swapping this into the denominator, we get:

$\frac{dy}{dx}=\frac{2\cos\theta\sin\theta}{1-\sin^{2}\theta-\sin^{2}\theta}$

$\frac{dy}{dx}=\frac{2\cos\theta\sin\theta}{1-2\sin^{2}\theta}$

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Example Question #1 : Polar Calculations

Convert the polar coordinate equation $r^2\cos^2\theta+r\sin\theta = 17$ into its rectangular equivalent, and simplify.

Possible Answers:

None of the other answers

x^2 + y^2 = 17

y = -x^2 +17

x = y^3 + 17

$y^2 + x^4 = \sqrt{17}$

Correct answer:

y = -x^2 +17

Explanation:

The polar to rectangular transformation equations are $x = r\cos\theta, y = r\sin\theta$ . By substituting these into our given equation we get x^2 + y = 17 .

Adding -x^2 to both sides, then we get y = -x^2 +17 .

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Example Question #1 : Polar Calculations

Convert the following polar coordinates of the form $\tiny (r,\theta )$ into Cartesian coordinates of the form $\tiny (x,y)$ :

$(-4,\frac{2\pi }{3})$

Possible Answers:

$(2\sqrt{3},-2)$

$(-4,3\sqrt{2})$

$(-4,-2\sqrt{3})$

$(2,-2\sqrt{3})$

$(4,-3\sqrt{2})$

Correct answer:

$(2,-2\sqrt{3})$

Explanation:

In order to convert the given polar coordinates into Cartesian coordinates, we must remember our formulas for x and y in terms of r and $\theta$ :

$x=rcos\theta$

$y=rsin\theta$

The problem tells us r and $\theta$ , so all we must do to convert these coordinates is plug them into the formulas above:

$x=rcos\theta=-4cos(\frac{2\pi }{3})=2$

$y=rsin\theta=-4sin(\frac{2\pi }{3})=-4(\frac{\sqrt{3}}{2})=-2\sqrt{3}$

So we can see from our conversion that the given polar coordinates are expressed as $(2,-2\sqrt{3})$ in Cartesian coordinates.

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Example Question #1 : Polar Calculations

Convert $\left ( 2,\frac{\pi}{4}\right )$ to Cartesian coordinates.

Possible Answers:

$\left ( \sqrt2,\sqrt2\right )$

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

$\left ( 2\sqrt2,2\sqrt2\right )$

$\left ( 2+\sqrt2,2+\sqrt2\right )$

$\left ( 2-\sqrt2,2-\sqrt2\right )$

Correct answer:

$\left ( \sqrt2,\sqrt2\right )$

Explanation:

The following formulas will convert polar coordinates to Cartesian coordinates.

$x=rcos\theta$

$y=rsin\theta$

We are given the polar coordinate, which is in $\left ( r,\theta\right )$ form. Plug the coordinate into the formulas and solve for x and y.

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

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

The Cartesian coordinate form is $\left ( \sqrt2, \sqrt2\right )$ .

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Example Question #321 : Parametric, Polar, And Vector

Given the Cartesian coordinate $\left ( -1,1\right )$ , what is $\theta$ in the polar form $\left ( r,\theta\right )$ ?

Possible Answers:

$\frac{7\pi}{6}$

$\frac{5\pi}{4}$

$\frac{3\pi}{4}$

$\frac{\pi}{4}$

$-\frac{\pi}{4}$

Correct answer:

$\frac{3\pi}{4}$

Explanation:

The formula to find theta in polar form is:

$\theta= tan^-^1\left(\frac{y}{x}\right)$

Plug in the Cartesian coordinates into the equation.

$\theta= tan^-^1\left(\frac{1}{-1}\right) = tan^-^1(-1) =-\frac{\pi}{4}$

However, this angle is located in the fourth quadrant and is not in the right quadrant. Add $\pi$ radians to get the correct angle since the Cartesian coordinate given is located in the second quadrant.

$-\frac{\pi}{4}+\pi = \frac{3\pi}{4}$

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Example Question #321 : Parametric, Polar, And Vector

Tom is scaling a mathematical mountain. The mountain's profile can be described by $\small \small \small -x^4+2x^3-x^2-x+3$ between $\small x=-1$ and $\small x\approx1.68$ . Tom climbs from $\small x=-1$ to the peak of the mountain. How far did he climb? You'll need an equation solver for certain parts of the problem. Round everything to the nearest hundredth.

Possible Answers:

$\small 3.15$

$\small 1.86$

$\small 7.18$

$\small 6.52$

$\small 3.33$

Correct answer:

$\small 3.33$

Explanation:

This is a two step problem. First step is to maximize the function to find the peak of the mountain. The next step is to use the arc length formula to find the distance he climbed.

To maximize, we'll take a derivative and set it equal to zero.

$\small \small \frac{\mathrm{d} }{\mathrm{d} x}(-x^4+2x^3-x^2-x+3)=-4x^3+6x^2-2x-1$ .

Setting this equal to zero, we get $\small \small x\approx-0.26$ .

The derivative is positive prior to this value and negative after, so it is a max. We now must take the arc length from $\small x=-1$ to $\small x\approx-0.26$ .

The formula for arc length is

$\small \small \int_{x=a}^{x=b}\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$ .

For this case, the integral becomes

$\small \small \int_{-1}^{-.26}\sqrt{1+(-4x^3+6x^2-2x-1)^2}dx$ .

This will give us $\small 3.33$ . No units were given in the problem, so leaving the answer unitless is acceptable.

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Example Question #2 : Polar Calculations

Find the length of the polar valued function $r=cos(\theta)$ from r=2 to r=5 .

Possible Answers:

-sin(5)+sin(2)

Correct answer:

Explanation:

Recall the formula for length in polar coordinates is given by

$L=\int_{a}^{b}\sqrt{r^2+\left(\frac{dr}{d \theta}\right)^2} d \theta$ .

We were given the formula

$r=cos(\theta)$ .

In our case, this translates to

$\\L=\int_{2}^{5}\sqrt{cos^2(\theta )+((-sin(\theta ))^2}d \theta\\ \\= \int_{2}^{5}\sqrt{1}d \theta\\ \\=3$ .

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Calculus 2 : Polar

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #822 : Calculus Ii

Example Question #315 : Parametric, Polar, And Vector

Example Question #311 : Parametric, Polar, And Vector

Example Question #21 : Derivatives Of Polar Form

Example Question #1 : Polar Calculations

Example Question #1 : Polar Calculations

Example Question #1 : Polar Calculations

Example Question #321 : Parametric, Polar, And Vector

Example Question #321 : Parametric, Polar, And Vector

Example Question #2 : Polar Calculations

All Calculus 2 Resources

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