Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Derivatives of Polar Form

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

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

Example Questions

← Previous 1 2 3 Next →

Example Question #11 : Derivatives Of Polar Form

Find the derivative of the following function:

$r=3\cos(\theta)+\theta^2$

Possible Answers:

$\frac{(-3\sin(\theta)+2\theta)\sin(\theta)+(3\cos(\theta)+\theta^2)(\cos(\theta))}{(-3\sin(\theta)+2\theta)(\cos(\theta))-(3\cos(\theta)+\theta^2)(\sin(\theta))}$

$\frac{(-3\sin(\theta)+2\theta)\sin(\theta)-(3\cos(\theta)+\theta^2)(\cos(\theta))}{(-3\sin(\theta)+2\theta)(\cos(\theta))-(3\cos(\theta)+\theta^2)(\sin(\theta))}$

$\frac{(-3\sin(\theta)+2\theta)\sin(\theta)-(3\cos(\theta)+\theta^2)(\cos(\theta))}{(-3\sin(\theta)+2\theta)(\cos(\theta))+(3\cos(\theta)-\theta^2)(\sin(\theta))}$

$\frac{(-3\sin(\theta)+2\theta)\cos(\theta)+(3\cos(\theta)+\theta^2)(\cos(\theta))}{(-3\sin(\theta)+2\theta)(\sin(\theta))-(3\cos(\theta)+\theta^2)(\sin(\theta))}$

Correct answer:

$\frac{(-3\sin(\theta)+2\theta)\sin(\theta)+(3\cos(\theta)+\theta^2)(\cos(\theta))}{(-3\sin(\theta)+2\theta)(\cos(\theta))-(3\cos(\theta)+\theta^2)(\sin(\theta))}$

Explanation:

The derivative of a polar function is given by

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)}$

First, we must find the derivative of the function, r:

$\frac{dr}{d\theta}=-3\sin(\theta)+2\theta$

We used the following rules to find the derivative:

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(\theta)=-\sin(\theta)$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Now, plug in the derivative and the original function r into the above formula:

$\frac{(-3\sin(\theta)+2\theta)\sin(\theta)+(3\cos(\theta)+\theta^2)(\cos(\theta))}{(-3\sin(\theta)+2\theta)(\cos(\theta))-(3\cos(\theta)+\theta^2)(\sin(\theta))}$

Report an Error

Example Question #12 : Derivatives Of Polar Form

Find the derivative of the following function:

$r=\theta+\tan(\theta)$

Possible Answers:

$\frac{(1+\sec^2(\theta))\cos(\theta)+(\theta+\tan(\theta))\sin(\theta)}{(1+\sec^2(\theta))\cos(\theta)-(\theta+\tan(\theta))\sin(\theta)}$

$\frac{(1+\sec^2(\theta))\sin(\theta)+(\theta+\tan(\theta))\cos(\theta)}{(1+\sec^2(\theta))\cos(\theta)+(\theta+\tan(\theta))\sin(\theta)}$

$\frac{(1+\sec^2(\theta))\sin(\theta)+(\theta+\tan(\theta))\cos(\theta)}{(1+\sec^2(\theta))\cos(\theta)-(\theta+\tan(\theta))\sin(\theta)}$

$\frac{(1+\sec^2(\theta))\sin(\theta)-(\theta+\tan(\theta))\cos(\theta)}{(1+\sec^2(\theta))\cos(\theta)-(\theta+\tan(\theta))\sin(\theta)}$

Correct answer:

$\frac{(1+\sec^2(\theta))\sin(\theta)+(\theta+\tan(\theta))\cos(\theta)}{(1+\sec^2(\theta))\cos(\theta)-(\theta+\tan(\theta))\sin(\theta)}$

Explanation:

The derivative (slope of the tangent line) of a polar function is given by the following formula:

$\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)}$

So, we must simply find $\frac{dr}{d\theta}$ and plug it into the above formula:

$\frac{dr}{d\theta}=1+\sec^2(\theta)$

The following rules were used to find the derivative:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{^{n-1}}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

Now, plug the given function and its derivative into the above formula to get our answer:

$\frac{(1+\sec^2(\theta))\sin(\theta)+(\theta+\tan(\theta))\cos(\theta)}{(1+\sec^2(\theta))\cos(\theta)-(\theta+\tan(\theta))\sin(\theta)}$

Report an Error

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:

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

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

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

Report an Error

Example Question #13 : Derivatives Of Polar Form

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

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

Report an Error

Example Question #4 : 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:

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

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

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

Report an Error

Example Question #11 : Derivatives Of Polar Form

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:

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

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

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

Report an Error

Example Question #15 : Derivatives Of Polar Form

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

Possible Answers:

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

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

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

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

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

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 #16 : 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:

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

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

Report an Error

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 #18 : Derivatives Of Polar Form

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:

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

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

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

Report an Error

← Previous 1 2 3 Next →

View Calculus 2 Tutors

Yoonsik
Certified Tutor

Seoul National University, Bachelor of Science, Physics. University of Pennsylvania, Doctor of Philosophy, Atomic and Molecul...

View Calculus 2 Tutors

Kurosh
Certified Tutor

University of tehran, Bachelor of Engineering, Electrical Engineering.

View Calculus 2 Tutors

Filiberto
Certified Tutor

University of Arizona, Bachelor of Science, Electrical Engineering.

All Calculus 2 Resources

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

Popular Subjects

Math Tutors in Seattle, Math Tutors in Miami, SAT Tutors in Boston, GRE Tutors in Atlanta, GMAT Tutors in Miami, Statistics Tutors in San Francisco-Bay Area, Statistics Tutors in Miami, Computer Science Tutors in Boston, Algebra Tutors in New York City, LSAT Tutors in Dallas Fort Worth

Popular Courses & Classes

GRE Courses & Classes in Miami, ACT Courses & Classes in Miami, SSAT Courses & Classes in Miami, Spanish Courses & Classes in New York City, LSAT Courses & Classes in Seattle, Spanish Courses & Classes in Chicago, ISEE Courses & Classes in Phoenix, ACT Courses & Classes in New York City, MCAT Courses & Classes in San Diego, LSAT Courses & Classes in Chicago

Popular Test Prep

ACT Test Prep in Atlanta, LSAT Test Prep in San Diego, MCAT Test Prep in Boston, ISEE Test Prep in Seattle, ISEE Test Prep in Phoenix, LSAT Test Prep in Boston, MCAT Test Prep in Philadelphia, MCAT Test Prep in Dallas Fort Worth, MCAT Test Prep in Denver, LSAT Test Prep in Seattle

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 2 : Derivatives of Polar Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Derivatives Of Polar Form

Example Question #12 : Derivatives Of Polar Form

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

Example Question #13 : Derivatives Of Polar Form

Example Question #4 : Derivatives Of Parametric, Polar, And Vector Functions

Example Question #11 : Derivatives Of Polar Form

Example Question #15 : Derivatives Of Polar Form

Example Question #16 : Derivatives Of Polar Form

Example Question #11 : Derivatives Of Polar Form

Example Question #18 : Derivatives Of Polar Form

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: