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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : Polar Coordinates

Write $r = 2 \cos \theta + \sin \theta$ in rectangular form

Possible Answers:

$(x+1)^2 + (y-\frac{1}{2})^2 = \frac{3}{4}$

$(x-1)^2 + (y-\frac{1}{2})^2 = 1 \frac{1}{4}$

$(x-1)^2 + (y-\frac{1}{2})^2 = 1\frac{1}{2}$

$(x-1)^2 + (y-\frac{1}{4})^2 = 1 \frac{1}{4}$

$(x+1)^2 + (y + \frac{1}{2})^2 = 1$

Correct answer:

$(x-1)^2 + (y-\frac{1}{2})^2 = 1 \frac{1}{4}$

Explanation:

multiply both sides by r:

$r^2 = 2 r \cos \theta + r \sin \theta$

Now we can make the substitutions r ^2 = x^2 + y^2 , $r \cos \theta = x$ , and $r \sin \theta = y$ :

x^2 + y^2 = 2x + y

If we subtract the terms on the right from both sides, we can complete the square twice and make this into a regular circle equation:

$x^2 - 2x \enspace \enspace + y^2 - y \enspace \enspace = 0$

To complete the square for x, add 1, since $(\frac{1}{2}(-2))^2= (-1)^ 2 =1$ .

To complete the square for y, add $\frac{1}{4}$ since $(\frac{1}{2}(-1))^2 = (\frac{1}{2})^2 = \frac{1}{4}$ .

$(x^2 - 2x + 1 ) + (y^2 - y + \frac{1}{4} ) =+ 1+\frac{1}{4}$

re-write the left side as binomials squared:

$(x-1)^2 + (y-\frac{1}{2})^2 = 1\frac{1}{4}$

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Example Question #22 : Convert Polar Equations To Rectangular Form And Vice Versa

Write the equation for $r = 8\sin \theta - 2 \cos \theta$ in rectangular form.

Possible Answers:

(x-1)^2 + (y-4)^2 = 17

(x+1)^2 + (y-4)^2 = 17

(x-2)^2 + (y+8)^2 = 68

(y-1)^2 + (x + 4)^2 = 17

(y+4)^2 + (x+1)^2 = 17

Correct answer:

(x+1)^2 + (y-4)^2 = 17

Explanation:

Multiply both sides by r:

$r^2 = 8r \sin \theta - 2 r \cos \theta$

Now we can substitute in r^2 = x^2 + y^2 , $x= r \cos \theta$ , and $y = r \sin \theta$ .

x^2 + y^2 = 8y-2x if we subtract both terms on the right from both sides, we can complete the square twice to get this into the normal form for a circle.

$x^2 + 2x \enspace \enspace + y^2 -8y \enspace \enspace =0$

To complete the square for x, add 1 to both sides, since $(\frac{1}{2}(2))^2 = (1)^2 = 1$

To complete the square for y, add 16 to both sides, since $(\frac{1}{2}(-8))^2 = (-4)^2 = 16$ :

(x^2 + 2x + 1 ) + (y^2 - 8y + 16 ) = +1 + 16 Re-write the left side as two binomials squared

(x+1)^2 + (y-4)^2 = 17

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Example Question #21 : Convert Polar Equations To Rectangular Form And Vice Versa

Write the equation for $r = \tan \theta$ in rectangular form

Possible Answers:

$y = \sqrt{ \frac{ -x^4 }{x^2 -1}}$

$y = \sqrt{\frac{ x^4}{ 1 + x^2 }}$

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

$y =\frac{1 + \sqrt{1-4x^4}}{2x}$

$y = \frac{x^4}{1 - x^2}$

Correct answer:

$y = \sqrt{ \frac{ -x^4 }{x^2 -1}}$

Explanation:

To convert, make the substitutions $r = \sqrt{x^2 + y^2 }$ and $\tan \theta = \frac{y}{x}$ :

$\sqrt{x^2 +y^2} = \frac{y}{x}$ square both sides

$x^2 + y^2 = \frac{y^2}{x^2}$ multiply both sides by x^2

x^2(x^2 + y^2 ) = y^2

x^4 + x^2 y^2 = y^2 subtract y^2 from both sides

x^4 + x^2 y^2 - y^2 = 0 subtract x^4 from both sides

x^2 y^2 - y^2 = -x^4 factor our y^2

y^2 (x^2 - 1 ) = -x^4 divide both sides by x^2 - 1

$y ^2 = \frac{ -x^4}{ x^2 - 1 }$ take the square root of both sides

$y = \sqrt{\frac{-x^4 }{ x^2 - 1 }}$

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Example Question #22 : Polar Coordinates

Convert the equation $r = \frac{ 8 \cos \theta }{1 + 7 \sin ^2 \theta }$ to rectangular form

Possible Answers:

$\frac{(x+4)^2}{16} + \frac{y^2 }{2} =1$

$\frac{(x+4)^2}{4} + \frac{y^2 }{2} =1$

$\frac{(x-4)^2}{16} + \frac{y^2 }{4} =1$

$\frac{(x+4)^2}{16} + \frac{y^2 }{4} =1$

$\frac{(x-4)^2}{16} + \frac{y^2 }{2} =1$

Correct answer:

$\frac{(x-4)^2}{16} + \frac{y^2 }{2} =1$

Explanation:

Multiply both sides by $1 + 7 \sin ^2 \theta$ :

$r(1 + 7 \sin ^2 \theta ) = 8 \cos \theta$ distribute

$r + 7 r \sin ^2 \theta = 8 \cos \theta$ multiply both sides by r

$r^2 + 7 r^2 \sin^2 \theta = 8 r \cos \theta$

convert to rectangular by making the substitutions x^2 + y^2 = r^2 , $y = r \sin \theta$ , and $x = r \cos \theta$ :

x^2 + y^2 + 7 y^2 = 8 x subtract 8x from both sides and combine like terms

x^2 - 8x + 8y^2 = 0 complete the square for x by adding 16 to both sides

x^2 - 8x + 16 + 8y^2 = 16

(x-4)^2 + 8y^2 = 16 divide both sides by 16

$\frac{(x-4)^2}{16 } + \frac{y^2}{2} = 1$

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Example Question #1621 : Pre Calculus

Convert (0,2) to polar coordinates.

Possible Answers:

(0,2)

$(\sqrt2,\pi)$

$\left(2,\frac{\pi}{2}\right)$

(2,0)

$\left(\sqrt2,\frac{\pi}{2}\right)$

Correct answer:

$\left(2,\frac{\pi}{2}\right)$

Explanation:

Write the Cartesian to polar conversion formulas.

r^2=x^2+y^2

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

Substitute the coordinate point to the equations and solve for $(r,\theta)$ .

r^2=x^2+y^2

$r=\sqrt{0^2+2^2}=2$

$\theta=tan^-^1 \left(\frac{y}{x}\right)=tan^-^1 \left(\frac{2}{0}\right)=\frac{\pi}{2}$

Since (0,2) is located in between the first and second quadrant, this is the correct angle.

Therefore, the answer is $\left(2,\frac{\pi}{2}\right)$ .

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Example Question #22 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert (-3,-3) to polar form.

Possible Answers:

$\left(3,\frac{\pi}{4}\right)$

$\left(3\sqrt2,\frac{5\pi}{4}\right)$

$\left(-3,-\frac{\pi}{4}\right)$

$\left(3,\frac{5\pi}{4}\right)$

$\left(3\sqrt2,\frac{\pi}{4}\right)$

Correct answer:

$\left(3\sqrt2,\frac{5\pi}{4}\right)$

Explanation:

Write the Cartesian to polar conversion formulas.

r^2=x^2+y^2

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

Substitute the coordinate point to the equations to find $(r,\theta)$ .

r^2=x^2+y^2

$r=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt2$

$\theta=tan^-^1 \left(\frac{y}{x}\right)=tan^-^1 \left(\frac{-3}{-3}\right)=\frac{\pi}{4}$

Since (-3,-3) is not located in between the first quadrant, this is not the correct angle. The correct location of this coordinate is in the third quadrant. Add $\pi$ radians to get the correct angle.

$\theta=\frac{\pi}{4}+\pi=\frac{5\pi}{4}$

Therefore, the answer is $\left(3\sqrt2,\frac{5\pi}{4}\right)$ .

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Example Question #23 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the rectangular equation to polar form:

(x+4)^2+y^2=16

Possible Answers:

$r(r+8\cos\theta)=0$

$8r\cos\theta+r+16=0$

r^2+8=0

$r(r+8r\cos\theta)=0$

Correct answer:

$r(r+8\cos\theta)=0$

Explanation:

Because $x=r\cos\theta$ and $y=r\sin\theta$ , substitute those values in the rectangular form.

(x+4)^2+y^2=16

$(r\cos\theta+4)^2+(r\sin\theta)^2=16$

Now, expand the equation.

$r^2\cos^2\theta+8r\cos\theta+16+r^2\sin^2\theta=16$

Subtract from both sides.

$r^2\cos^2\theta+8r\cos\theta+r^2\sin^2\theta=0$

$r^2(\cos^2\theta+\sin^2\theta)+8r\cos\theta=0$

Recall the trigonometric identity $\sin^2\theta+\cos^2\theta=1$

$r^2+8r\cos\theta=0$

Factor the equation.

$r(r+8\cos\theta)=0$

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Example Question #31 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the rectangular equation into polar form.

y=3x+2

Possible Answers:

$r=\sin\theta-3\cos\theta$

$r=\frac{\sin\theta-3\cos\theta}{2}$

$r=2(\sin\theta-3\cos\theta)$

$r=\frac{2}{\sin\theta-3\cos\theta}$

Correct answer:

$r=\frac{2}{\sin\theta-3\cos\theta}$

Explanation:

Recall that $x=r\cos\theta$ and $y=r\sin\theta$ .

Substitute these values into the equation.

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

Now, manipulate the equation so that the terms with are on the same side.

$r\sin\theta-3r\cos\theta=2$

Factor out the .

$r(\sin\theta-3\cos\theta)=2$

Divide both sides by $\sin\theta-3\cos\theta$ .

$r=\frac{2}{\sin\theta-3\cos\theta}$

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Example Question #32 : Convert Polar Equations To Rectangular Form And Vice Versa

Conver the rectanglar equation into polar form.

y=-2x-5

Possible Answers:

$r=-5(\sin\theta+2\cos\theta)$

$r=-\frac{\sin\theta+2\cos\theta}{5}$

$r=-\frac{5}{\sin\theta+2\cos\theta}$

$r=-2\cos\theta-\sin\theta$

Correct answer:

$r=-\frac{5}{\sin\theta+2\cos\theta}$

Explanation:

Recall that $x=r\cos\theta$ and $y=r\sin\theta$ .

Substitute these values into the equation.

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

Now, manipulate the equation so that the terms with are on the same side.

$r\sin\theta+2r\cos\theta=-5$

Factor out the .

$r(\sin\theta+2\cos\theta)=-5$

Divide both sides by $\sin\theta+2\cos\theta$ .

$r=-\frac{5}{\sin\theta+2\cos\theta}$

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Example Question #101 : Polar Coordinates And Complex Numbers

Convert the rectangular equation to polar form.

x=7y+2

Possible Answers:

$r=2(\cos\theta-7\sin\theta)$

$r=\frac{2}{\cos\theta-7\sin\theta}$

$r=\frac{\cos\theta-7\sin\theta}{2}$

$r=\frac{2}{\sin\theta-7\cos\theta}$

Correct answer:

$r=\frac{2}{\cos\theta-7\sin\theta}$

Explanation:

Recall that $x=r\cos\theta$ and $y=r\sin\theta$ .

Substitute these values into the equation.

$r\cos\theta=7(r\sin\theta)+2$

Now, manipulate the equation so that the terms with are on the same side.

$r\cos\theta-7r\sin\theta=2$

Factor out the .

$r(\cos\theta-7\sin\theta)=2$

Divide both sides by $\cos\theta-7\sin\theta$ .

$r=\frac{2}{\cos\theta-7\sin\theta}$

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #21 : Polar Coordinates

Example Question #22 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #21 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #22 : Polar Coordinates

Example Question #1621 : Pre Calculus

Example Question #22 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #23 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #31 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #32 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #101 : Polar Coordinates And Complex Numbers

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