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Calculus 2 : Calculus II

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

Example Questions

Example Question #811 : Calculus Ii

Find the derivative of the following function:

$r=\theta^2+1$

Possible Answers:

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

$2\theta$

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

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

Correct answer:

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

Explanation:

The derivative of a polar function is given by the following:

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

First, we must find $\frac{dr}{d\theta}=2\theta$

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Finally, plug in the derivative we just found along with r, the function given, into the above formula:

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

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Example Question #2 : Derivatives Of Polar Form

Find the derivative of the following function:

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

Possible Answers:

$-2\cos(\theta)\sin(\theta)$

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

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

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

Correct answer:

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

Explanation:

The derivative of a polar function is given by the following:

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

First, we must find $\frac{dr}{d\theta}=-2\cos(\theta)\sin(\theta)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x)) \cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Finally, plug in the above derivative and our original function into the above formula:

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

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Example Question #2 : Derivatives Of Polar Form

Find the derivative of the following function:

$r=3\theta+e^\theta$

Possible Answers:

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

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

$3+e^\theta$

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

Correct answer:

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

Explanation:

The derivative of a polar function is given by the following:

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

First, we must find $\frac{dr}{d\theta}=3+e^\theta$

We found the derivative using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x}e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Finally, we plug in the above derivative and the original function into the above formula:

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

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Example Question #7 : Derivatives Of Polar Form

Find the derivative of the function:

$r=5\cos(2\theta)$

Possible Answers:

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

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

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

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

Correct answer:

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

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:

$\frac{dr}{d\theta}=-10\sin(2\theta)$

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Now, using the derivative we just found and our original function in the above formula, we can write out the derivative of the function in terms of x and y:

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

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Example Question #301 : Parametric, Polar, And Vector

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:

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

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Example Question #302 : Parametric, Polar, And Vector

Find the derivative of the following function:

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

Possible Answers:

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

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

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

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

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:

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

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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 #811 : Calculus Ii

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 #303 : Parametric, Polar, And Vector

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #811 : Calculus Ii

Example Question #2 : Derivatives Of Polar Form

Example Question #2 : Derivatives Of Polar Form

Example Question #7 : Derivatives Of Polar Form

Example Question #301 : Parametric, Polar, And Vector

Example Question #302 : Parametric, Polar, And Vector

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

Example Question #142 : Polar

Example Question #811 : Calculus Ii

Example Question #303 : Parametric, Polar, And Vector

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