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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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

Example Questions

Example Question #841 : Calculus Ii

Possible Answers:

$r=\sqrt{}2$

$\Theta=\frac{4}{\Pi}$

r=2

$\Theta=\frac{5\Pi}{4}$

$r=\sqrt{2}$

$\Theta=\frac{\Pi}{4}$

$r=\sqrt{2}$

$\Theta=\frac{5\Pi}{4}$

r=2

$\Theta=\frac{\Pi}{4}$

Correct answer:

$r=\sqrt{2}$

$\Theta=\frac{5\Pi}{4}$

Explanation:

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Example Question #12 : Polar Calculations

Determine how many points of intersection exist for the curves

$r=cos(2\Theta)$

and

$r = \frac{1}{2}$ .

Possible Answers:

Correct answer:

Explanation:

Solving the equations $r=cos(2\Theta )$ and $r=\frac{1}{2}$ yields $cos(2\Theta)=\frac{1}{2}$ .

Hence,

$2\Theta=\frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}.$

Therefore, the values of $\Theta$ between and $2\pi$ that satisfy both equations are:

$\Theta=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}.$

From this, it can be deduced that there are four points of intersection between the given curves:

$(\frac{1}{2},\frac{\pi}{6}),(\frac{1}{2},\frac{5\pi}{6}),(\frac{1}{2},\frac{7\pi}{6}),(\frac{1}{2},\frac{11\pi}{6}).$

However, an identical graph to $r=\frac{1}{2}$ in polar coordinates is $r=-\frac{1}{2}$ , since these two equations describe the same circle with a radius $\frac{1}{2}$ units long. Therefore, the equations $r=cos(2\Theta)$ and $r=-\frac{1}{2}$ must also be solved to yield the remaining points of intersection:

$cos(2\Theta)=-\frac{1}{2}$ ,

$\Rightarrow 2\Theta=\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}.$

$\Rightarrow \Theta=\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}.$

From this, it can be deduced that there are four other points of intersection between the given curves:

$(-\frac{1}{2},\frac{2\pi}{3}),(-\frac{1}{2},\frac{2\pi}{3}),(-\frac{1}{2},\frac{4\pi}{3}),(-\frac{1}{2},\frac{5\pi}{3}).$

Hence, there are eight total points of intersection between the curves $r=cos(2\Theta)$ and $r=\frac{1}{2}$ .

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Example Question #13 : Polar Calculations

Convert $r=-3cos(\theta)$ to Cartesian coordinates

Possible Answers:

x+y=-3x

x^2+y^2=-3y

x+y=-3y

x^2+y^2=-3x

x^2+y^2=y

Correct answer:

x^2+y^2=-3x

Explanation:

we are given $r=-3cos(\theta)$ and we know that

$x=rcos(\theta )$

$y=rsin(\theta )$

r^2=x^2+y^2

we have r and 3cos(theta). Multiplying each side of the equation by r would give us

$r^2=-3rcos(\theta )$

substitute out the parts we know from the formulas above

x^2+y^2=-3x

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Example Question #161 : Polar

Convert the polar point $(-5, 5 \pi/6)$ into cartesian coordinates.

Possible Answers:

$(-5, 5 \pi/6)$

$( 5\sqrt3/4,5\sqrt3/4)$

$(5\sqrt3/2, -5/2)$

$(5/2, 5\sqrt3/2)$

Correct answer:

$(5\sqrt3/2, -5/2)$

Explanation:

Cartesian coordinates are written in the form $(r,\theta)$ .

In this problem, r=-5 and $\theta=5 \pi/6$ .

Using the conversion formulas $x=rcos\theta$ and $y=rsin\theta$ ,

$x=-5*cos(5\pi/6)=-5(-\sqrt3/2)=5\sqrt3/2$

$y=-5*sin(5\pi/6)=-5(1/2)=-5/2$

The cartesian point is $(5\sqrt3/2, -5/2)$ .

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Example Question #162 : Polar

Convert the polar point $(2, \pi/3)$ into cartesian coordinates.

Possible Answers:

$(\sqrt3,1)$

(1,1)

$(1,\sqrt3)$

$( 5\sqrt3/4,5\sqrt3/4)$

Correct answer:

$(1,\sqrt3)$

Explanation:

Cartesian coordinates are written in the form $(r,\theta)$ .

In this problem, r=2 and $\theta= \pi/3$ .

Using the conversion formulas $x=rcos\theta$ and $y=rsin\theta$ ,

$x=2*cos(\pi/3)=2(1/2)=1$

$y=2*sin(\pi/3)=2(\sqrt3/2)=\sqrt3$

The cartesian point is $(1,\sqrt3)$ .

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Example Question #841 : Calculus Ii

Convert the polar point $(-8, \pi)$ into cartesian coordinates.

Possible Answers:

$(-8, \pi)$

(8,0)

(8,1)

$(1,\sqrt3)$

Correct answer:

(8,0)

Explanation:

Cartesian coordinates are written in the form $(r,\theta)$ .

In this problem, r=-8 and $\theta=\pi$ .

Using the conversion formulas $x=rcos\theta$ and $y=rsin\theta$ ,

$x=-8*cos(\pi)=-8(-1)=8$

$y=-8*sin(\pi)=-8(0)=0$

The cartesian point is (8,0) .

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Example Question #17 : Polar Calculations

Convert the cartesian point (1,1) into polar coordinates.

Possible Answers:

$(-8, \pi)$

(-1,1)

$(\sqrt3/2, \pi/6)$

$(\sqrt2, \pi/4)$

Correct answer:

$(\sqrt2, \pi/4)$

Explanation:

Cartesian coordinates are written in the form (x,y) .

In this problem, x=1 and y=1 .

Using the conversion formulas $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}\left ( y/x\right )$ ,

$r=\sqrt{x^2+y^2}=\sqrt{1^2+1^2}=\sqrt2$

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

The polar point is $(\sqrt2, \pi/4)$ .

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Example Question #331 : Parametric, Polar, And Vector

Convert the cartesian point (-3,4) into polar coordinates.

Possible Answers:

(5,5.356)

(-3,4)

$(\sqrt3/2, \pi/6)$

(-4,2.584)

Correct answer:

(5,5.356)

Explanation:

Cartesian coordinates are written in the form (x,y) .

In this problem, x=-3 and y=4 .

Using the conversion formulas $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}\left ( y/x\right )$ ,

$r=\sqrt{x^2+y^2}=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

$\theta=tan^{-1}\left ( y/x\right )=tan^{-1}\left ( -4/3\right )=-0.927=5.356$

The polar point is (5,5.356) .

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Example Question #19 : Polar Calculations

Convert the cartesian point (5, 12) into polar coordinates.

Possible Answers:

(12,3.746)

(5, 12)

(13,1.176)

(-3,4)

Correct answer:

(13,1.176)

Explanation:

Cartesian coordinates are written in the form (x,y) .

In this problem, x=5 and y=12 .

Using the conversion formulas $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}\left ( y/x\right )$ ,

$r=\sqrt{x^2+y^2}=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$

$\theta=tan^{-1}\left ( y/x\right )=tan^{-1}\left ( 12/5\right )=1.176$

The polar point is (13,1.176) .

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Example Question #20 : Polar Calculations

Convert xy+7x^2=1+y into polar coordinates.

Possible Answers:

$r^2\left (sin\theta cos\theta+7sin^2\theta \right )-rcos\theta=7$

$rsin\theta=4$

$r^2\left (sin\theta cos\theta+7cos^2\theta \right )-rsin\theta=1$

$r^2\left (sin\theta cos\theta+7cos^2\theta \right )=-1$

Correct answer:

$r^2\left (sin\theta cos\theta+7cos^2\theta \right )-rsin\theta=1$

Explanation:

Substituting the conversion formulas $x=rcos\theta$ and $y=rsin\theta$ into the cartesian equation,

xy+7x^2=1+y

we get

$(rcos\theta)(rsin\theta)+7(rcos\theta)^2=1+(rsin\theta)$

$r^2sin\theta cos\theta+7r^2cos^2\theta=1+rsin\theta$

$r^2\left (sin\theta cos\theta+7cos^2\theta \right )-rsin\theta=1$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #841 : Calculus Ii

Example Question #12 : Polar Calculations

Example Question #13 : Polar Calculations

Example Question #161 : Polar

Example Question #162 : Polar

Example Question #841 : Calculus Ii

Example Question #17 : Polar Calculations

Example Question #331 : Parametric, Polar, And Vector

Example Question #19 : Polar Calculations

Example Question #20 : Polar Calculations

All Calculus 2 Resources

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