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Calculus 2 : Polar

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Example Questions

Example Question #31 : Parametric, Polar, And Vector Functions

What is the derivative of $r=4sin\theta+2$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=4sin\theta+2$ , we must first find the derivative of with respect to $\theta$ as follows:

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

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:

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

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

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

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

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:

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

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

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

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Example Question #142 : Polar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=3-5sin\theta$ , we must first find the derivative of with respect to $\theta$ as follows:

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

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:

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

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

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

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

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can simplify the denominator to be

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

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Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

What is the derivative of $r=7sin\theta-6$ ?

Possible Answers:

$\frac{14cos\theta\sin\theta+6\cos\theta}{7-14\sin^{2}\theta+6\sin\theta}$

$\frac{14cos\theta\sin\theta-6\cos\theta}{7-14\sin^{2}\theta+6\sin\theta}$

$\frac{14cos\theta\sin\theta-6\cos\theta}{7+14\sin^{2}\theta+6\sin\theta}$

$-\frac{14cos\theta\sin\theta-6\cos\theta}{7-14\sin^{2}\theta+6\sin\theta}$

$\frac{14cos\theta\sin\theta-6\cos\theta}{7-14\sin^{2}\theta-6\sin\theta}$

Correct answer:

$\frac{14cos\theta\sin\theta-6\cos\theta}{7-14\sin^{2}\theta+6\sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=7sin\theta-6$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=7cos\theta$

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:

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

$\frac{dy}{dx}=\frac{7cos\theta\sin\theta+(7sin\theta-6)\cos\theta}{7cos\theta\cos\theta-(7sin\theta-6)\sin\theta}$

$\frac{dy}{dx}=\frac{7cos\theta\sin\theta+7sin\theta\cos\theta-6\\cos\theta}{7cos\theta\cos\theta-7sin\theta\sin\theta+6\sin\theta}$

$\frac{dy}{dx}=\frac{14cos\theta\sin\theta-6\cos\theta}{7(cos^{2}\theta-sin^{2}\theta)+6\sin\theta}$

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:

$\frac{dy}{dx}=\frac{14cos\theta\sin\theta-6\cos\theta}{7((1-\sin^{2}\theta)-sin^{2}\theta)+6\sin\theta}$

$\frac{dy}{dx}=\frac{14cos\theta\sin\theta-6\cos\theta}{7(1-2\sin^{2}\theta)+6\sin\theta}$

$\frac{dy}{dx}=\frac{14cos\theta\sin\theta-6\cos\theta}{7-14\sin^{2}\theta+6\sin\theta}$

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Example Question #811 : Calculus Ii

What is the derivative of $r=7\cos\theta+6$ ?

Possible Answers:

$\frac{14\sin^{2}\theta+7+6\cos\theta}{-14\sin\theta\cos\theta-6\sin\theta}$

$\frac{-14\sin^{2}\theta+7+6\cos\theta}{14\sin\theta\cos\theta-6\sin\theta}$

$\frac{-14\sin^{2}\theta+7+6\cos\theta}{-14\sin\theta\cos\theta-6\sin\theta}$

$\frac{-14\sin^{2}\theta+7-6\cos\theta}{-14\sin\theta\cos\theta-6\sin\theta}$

$\frac{-14\sin^{2}\theta-7+6\cos\theta}{-14\sin\theta\cos\theta-6\sin\theta}$

Correct answer:

$\frac{-14\sin^{2}\theta+7+6\cos\theta}{-14\sin\theta\cos\theta-6\sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=7\cos\theta+6$ , 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:

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

$\frac{dy}{dx}=\frac{(-7\sin\theta)sin\theta+(7\cos\theta+6)\cos\theta}{\(-7\sin\theta)cos\theta-(7\cos\theta+6)sin\theta}$

$\frac{dy}{dx}=\frac{-7\sin^{2}\theta+7\cos^{2}\theta+6\cos\theta}{-7\sin\theta\cos\theta-7\cos\theta\sin\theta-6\sin\theta}$

$\frac{dy}{dx}=\frac{-7(\sin^{2}\theta-\cos^{2}\theta)+6\cos\theta}{-14\sin\theta\cos\theta-6\sin\theta}$

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 numerator, we get:

$\frac{dy}{dx}=\frac{-7(\sin^{2}\theta-(1-\sin^{2}\theta))+6\cos\theta}{-14\sin\theta\cos\theta-6\sin\theta}$

$\frac{dy}{dx}=\frac{-7(2\sin^{2}\theta-1)+6\cos\theta}{-14\sin\theta\cos\theta-6\sin\theta}$

$\frac{dy}{dx}=\frac{-14\sin^{2}\theta+7+6\cos\theta}{-14\sin\theta\cos\theta-6\sin\theta}$

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Example Question #144 : Polar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=3+5sin\theta$ , we must first find the derivative of with respect to $\theta$ as follows:

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

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:

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

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

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

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

Report an Error

Example Question #11 : Derivatives Of Polar Form

What is the derivative of $r=sin\theta-12$ ?

Possible Answers:

$\frac{2\cos\theta\sin\theta-12\cos\theta}{1+2\sin^{2}\theta+12\sin\theta}$

$\frac{2\cos\theta\sin\theta-12\cos\theta}{1-2\sin^{2}\theta-12\sin\theta}$

$\frac{2\cos\theta\sin\theta-12\cos\theta}{1-2\sin^{2}\theta+12\sin\theta}$

$\frac{2\cos\theta\sin\theta+12\cos\theta}{1-2\sin^{2}\theta+12\sin\theta}$

$\frac{2\cos\theta\sin\theta+12\cos\theta}{1+2\sin^{2}\theta+12\sin\theta}$

Correct answer:

$\frac{2\cos\theta\sin\theta-12\cos\theta}{1-2\sin^{2}\theta+12\sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=sin\theta-12$ , 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:

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

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

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

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

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:

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

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

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Example Question #11 : Derivatives Of Polar Form

What is the derivative of $r=7+2\sin\theta$ ?

Possible Answers:

$\frac{4\cos\theta\sin\theta-7\cos\theta}{2-4\sin^{2}\theta-7\sin\theta}$

$\frac{4\cos\theta\sin\theta-7\cos\theta}{2+4\sin^{2}\theta-7\sin\theta}$

$\frac{4\cos\theta\sin\theta+7\cos\theta}{2-4\sin^{2}\theta-7\sin\theta}$

$\frac{4\cos\theta\sin\theta+7\cos\theta}{2+4\sin^{2}\theta-7\sin\theta}$

$\frac{4\cos\theta\sin\theta+7\cos\theta}{2-4\sin^{2}\theta+7\sin\theta}$

Correct answer:

$\frac{4\cos\theta\sin\theta+7\cos\theta}{2-4\sin^{2}\theta-7\sin\theta}$

Explanation:

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=7+2\sin\theta$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=2\cos\theta$

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:

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

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

$\frac{dy}{dx}=\frac{2\cos\theta\sin\theta+7\cos\theta+2\sin\theta\cos\theta}{2\cos\theta\cos\theta-7\sin\theta-2\sin\theta\sin\theta}$

$\frac{dy}{dx}=\frac{4\cos\theta\sin\theta+7\cos\theta}{2\cos^{2}\theta-7\sin\theta-2\sin^{2}\theta}$

$\frac{dy}{dx}=\frac{4\cos\theta\sin\theta+7\cos\theta}{2(\cos^{2}\theta-\sin^{2}\theta)-7\sin\theta}$

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:

$\frac{dy}{dx}=\frac{4\cos\theta\sin\theta+7\cos\theta}{2(1-\sin^{2}\theta-\sin^{2}\theta)-7\sin\theta}$

$\frac{dy}{dx}=\frac{4\cos\theta\sin\theta+7\cos\theta}{2(1-2\sin^{2}\theta)-7\sin\theta}$

$\frac{dy}{dx}=\frac{4\cos\theta\sin\theta+7\cos\theta}{2-4\sin^{2}\theta-7\sin\theta}$

Report an Error

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 #141 : Polar

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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Calculus 2 : Polar

Study concepts, example questions & explanations for Calculus 2

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Example Question #31 : Parametric, Polar, And Vector Functions

Example Question #142 : Polar

Example Question #1 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #811 : Calculus Ii

Example Question #144 : Polar

Example Question #11 : Derivatives Of Polar Form

Example Question #11 : Derivatives Of Polar Form

Example Question #144 : Polar

Example Question #141 : Polar

Example Question #313 : Parametric, Polar, And Vector

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