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Calculus 2 : Parametric, Polar, and Vector

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Example Questions

Example Question #144 : Polar

What is the derivative of $r=3-5\sin\theta$ ?

Possible Answers:

$\frac{-10\cos\theta\sin\theta+3\cos\theta}{-5-10\sin^{2}\theta+3\sin\theta}$

$\frac{-10\cos\theta\sin\theta+3\cos\theta}{-5+10\sin^{2}\theta-3\sin\theta}$

$\frac{-10\cos\theta\sin\theta+3\cos\theta}{5-10\sin^{2}\theta-3\sin\theta}$

$\frac{-10\cos\theta\sin\theta-3\cos\theta}{-5-10\sin^{2}\theta-3\sin\theta}$

$\frac{-10\cos\theta\sin\theta+3\cos\theta}{-5-10\sin^{2}\theta-3\sin\theta}$

Correct answer:

$\frac{-10\cos\theta\sin\theta+3\cos\theta}{-5-10\sin^{2}\theta-3\sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=3-5\sin\theta$ , 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:

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

$\frac{dy}{dx}=\frac{-5\cos\theta\sin\theta+(3-5\sin\theta)\cos\theta}{-5\cos\theta\cos\theta-(3-5\sin\theta)\sin\theta}$

$\frac{dy}{dx}=\frac{-5\cos\theta\sin\theta+3\cos\theta-5\sin\theta\cos\theta}{-5\cos\theta\cos\theta-3\sin\theta+5\sin\theta\sin\theta}$

$\frac{dy}{dx}=\frac{-10\cos\theta\sin\theta+3\cos\theta}{-5(\cos^{2}\theta-\sin^{2}\theta)-3\sin\theta}$

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:

$\frac{dy}{dx}=\frac{-10\cos\theta\sin\theta+3\cos\theta}{-5(1-\sin^{2}\theta-\sin^{2}\theta)-3\sin\theta}$

$\frac{dy}{dx}=\frac{-10\cos\theta\sin\theta+3\cos\theta}{-5(1-2\sin^{2}\theta)-3\sin\theta}$

$\frac{dy}{dx}=\frac{-10\cos\theta\sin\theta+3\cos\theta}{-5-10\sin^{2}\theta-3\sin\theta}$

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Example Question #821 : Calculus Ii

What is the derivative of $r=4\sin\theta+5$ ?

Possible Answers:

$\frac{8\cos\theta\sin\theta-5\cos\theta}{4-8\sin^{2}\theta-5\sin\theta}$

$\frac{8\cos\theta\sin\theta+5\cos\theta}{4+8\sin^{2}\theta-5\sin\theta}$

$\frac{8\cos\theta\sin\theta+5\cos\theta}{4-8\sin^{2}\theta-5\sin\theta}$

$\frac{8\cos\theta\sin\theta+5\cos\theta}{4+8\sin^{2}\theta+5\sin\theta}$

$\frac{8\cos\theta\sin\theta+5\cos\theta}{4-8\sin^{2}\theta+5\sin\theta}$

Correct answer:

$\frac{8\cos\theta\sin\theta+5\cos\theta}{4-8\sin^{2}\theta-5\sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=4\sin\theta+5$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=4\cos\theta$

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:

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

$\frac{dy}{dx}=\frac{4\cos\theta\sin\theta+(4\sin\theta+5)\cos\theta}{4\cos\theta\cos\theta-(4\sin\theta+5)\sin\theta}$

$\frac{dy}{dx}=\frac{4\cos\theta\sin\theta+4\sin\theta\cos\theta+5\cos\theta}{4\cos\theta\cos\theta-4\sin\theta\sin\theta-5\sin\theta}$

$\frac{dy}{dx}=\frac{8\cos\theta\sin\theta+5\cos\theta}{4(\cos^{2}\theta-\sin^{2}\theta)-5\sin\theta}$

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:

$\frac{dy}{dx}=\frac{8\cos\theta\sin\theta+5\cos\theta}{4(1-\sin^{2}\theta-\sin^{2}\theta)-5\sin\theta}$

$\frac{dy}{dx}=\frac{8\cos\theta\sin\theta+5\cos\theta}{4(1-2\sin^{2}\theta)-5\sin\theta}$

$\frac{dy}{dx}=\frac{8\cos\theta\sin\theta+5\cos\theta}{4-8\sin^{2}\theta-5\sin\theta}$

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Example Question #313 : Parametric, Polar, And Vector

What is the derivative of $r=8-3\sin\theta$ ?

Possible Answers:

$\frac{8\cos\theta+6\sin\theta\cos\theta}{-3+6\sin^{2}\theta-8\sin\theta}$

$\frac{8\cos\theta-6\sin\theta\cos\theta}{-3+6\sin^{2}\theta+8\sin\theta}$

$\frac{8\cos\theta-6\sin\theta\cos\theta}{3+6\sin^{2}\theta-8\sin\theta}$

$\frac{8\cos\theta-6\sin\theta\cos\theta}{-3+6\sin^{2}\theta-8\sin\theta}$

$\frac{8\cos\theta-6\sin\theta\cos\theta}{-3-6\sin^{2}\theta-8\sin\theta}$

Correct answer:

$\frac{8\cos\theta-6\sin\theta\cos\theta}{-3+6\sin^{2}\theta-8\sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=8-3\sin\theta$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=-3\cos\theta$

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:

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

$\frac{dy}{dx}=\frac{(-3\cos\theta)\sin\theta+(8-3\sin\theta)\cos\theta}{(-3\cos\theta)\cos\theta-(8-3\sin\theta)\sin\theta}$

$\frac{dy}{dx}=\frac{-3\cos\theta\sin\theta+8\cos\theta-3\sin\theta\cos\theta}{-3\cos^{2}\theta-8\sin\theta+3\sin^{2}\theta}$

$\frac{dy}{dx}=\frac{8\cos\theta-6\sin\theta\cos\theta}{-3(\cos^{2}\theta-\sin^{2}\theta)-8\sin\theta}$

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:

$\frac{dy}{dx}=\frac{8\cos\theta-6\sin\theta\cos\theta}{-3(1-\sin^{2}\theta-\sin^{2}\theta)-8\sin\theta}$

$\frac{dy}{dx}=\frac{8\cos\theta-6\sin\theta\cos\theta}{-3(1-2\sin^{2}\theta)-8\sin\theta}$

$\frac{dy}{dx}=\frac{8\cos\theta-6\sin\theta\cos\theta}{-3+6\sin^{2}\theta-8\sin\theta}$

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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:

x^2 + y^2 = 17

x = y^3 + 17

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

None of the other answers

y = -x^2 +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:

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

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

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

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

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

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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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #144 : Polar

Example Question #821 : Calculus Ii

Example Question #313 : Parametric, Polar, And Vector

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

All Calculus 2 Resources

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