Math / Calculus

Question-Multiple Choice Questions a. 5 b. 10
2. The equation of the leminscate' $r^{2}=4 \cos 2 \theta$ is equivalent to the rectangular equation : a. $\left(x^{2}+y^{2}\right)^{2}=4\left(x^{2}-y^{2}\right)$ b. $\left(x^{2}+y^{2}\right)^{2}=4$ c. $\left(x^{2}+y^{2}\right)=4\left(x^{2}-y^{2}\right)$ d. $\left(x^{2}+y^{2}\right)^{2}=4\left(x^{2}-y^{2}\right)^{2}$
3. The curve of $r=8 \sin 3 \theta$ is a a. 6 -petals rose b. 3 - petals rose c. 16 - petals rose d. 8 - petals rose $\int_{0}^{\frac{1}{2}} \frac{d x}{\sqrt{\left(1-x^{2}\right)^{3}}}=$ a. $\frac{1}{\sqrt{3}}$ b. $\frac{1}{\sqrt{5}}$ c. $\frac{1}{\sqrt{15}}$ d. $\frac{1}{\sqrt{2}}$
5. $\int \frac{6 x+4}{x^{3}+4 x} d x=$ a. $2 \ln |x|+\ln \left(x^{2}+4\right)+6 \tan ^{-1}\left(\frac{x}{2}\right)+c$ c. $\ln |x|-\frac{1}{2} \ln \left(x^{2}+4\right)+3 \tan ^{-1}\left(\frac{x}{2}\right)+c$ b. $\ln |x|+\frac{1}{2} \ln \left(x^{2}+4\right)+3 \tan ^{-1}\left(\frac{x}{2}\right)+c$ d. $2 \ln |x|-\ln \left(x^{2}+4\right)+6 \tan ^{-1}\left(\frac{x}{2}\right)+c$
6. If the sum of the series $\sum_{k=1}^{\infty} a_{k}$ is 10 , then $\lim _{n \rightarrow \infty}\left(a_{n}+\frac{6 n+1}{2 n+3}\right)=$ a. 13 b. 3 c. 4 d. 14
7. The sum of the series $\sum_{k=2}^{\infty} \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k^{2}+k}}$ is a. $\frac{1}{\sqrt{3}}$ b. $\frac{1}{\sqrt{2}}$ c. 1 d. not exists(divergent)

Studdy Solution

STEP 1

1. For the conversion of polar to rectangular coordinates, we'll use the equations $x = r \cos \theta$ and $y = r \sin \theta$ .
2. The properties of trigonometric identities and their integrals might be necessary.
3. Knowledge of complex numbers and their properties might be required.
4. Understanding of series convergence and divergence.
5. Knowledge of the properties of parametric equations and their graphs.

STEP 2

1. Solve the problem of converting the polar equation of the lemniscate to a rectangular equation.
2. Determine the curve represented by the polar equation $r = 8 \sin 3\theta$ .
3. Evaluate the integral $\int_{0}^{\frac{1}{2}} \frac{d x}{\sqrt{\left(1-x^{2}\right)^{3}}}$ .
4. Evaluate the integral $\int \frac{6 x+4}{x^{3}+4 x} d x$ .
5. Determine the limit $\lim _{n \rightarrow \infty}\left(a_{n}+\frac{6 n+1}{2 n+3}\right)$ given the sum of the series $\sum_{k=1}^{\infty} a_{k}$ .
6. Find the sum of the series $\sum_{k=2}^{\infty} \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k^{2}+k}}$ .

STEP 3

Start with the given polar equation for the lemniscate: $r^2 = 4 \cos 2\theta$ .

STEP 4

Use the double-angle identity for cosine: $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ .

STEP 5

Substitute $x = r \cos \theta$ and $y = r \sin \theta$ into the equation. $r^2 = x^2 + y^2$ and $\cos 2\theta = \frac{x^2 - y^2}{x^2 + y^2}$ .

STEP 6

Combine the equations to get $\left(x^2 + y^2\right)^2 = 4(x^2 - y^2)$ . Thus, the correct answer is: a. $\left(x^2 + y^2\right)^2 = 4(x^2 - y^2)$ .

STEP 7

Analyze the curve given by the polar equation $r = 8 \sin 3 \theta$ .

STEP 8

Recognize that the equation $r = a \sin n \theta$ represents a rose curve with $n$ petals if $n$ is odd, otherwise $2n$ petals. Here, $n = 3$ , so it is a 3-petals rose. Thus, the correct answer is: b. 3-petals rose.

STEP 9

Evaluate the integral $\int_{0}^{\frac{1}{2}} \frac{d x}{\sqrt{\left(1-x^{2}\right)^{3}}}$ by substituting $x = \sin \theta$ , $dx = \cos \theta \, d\theta$ .

STEP 10

Rewrite the integral in terms of $\theta$ : $\int_{0}^{\frac{\pi}{6}} \frac{\cos \theta \, d\theta}{\left(\cos^2 \theta\right)^{3/2}} = \int_{0}^{\frac{\pi}{6}} \sec \theta \, d\theta$ .

STEP 11

Solve the integral: $\int_{0}^{\frac{\pi}{6}} \sec \theta \, d\theta = \left[\ln |\sec \theta + \tan \theta|\right]_{0}^{\frac{\pi}{6}}$ .

STEP 12

Evaluate the expression: $\ln |\sec (\frac{\pi}{6}) + \tan (\frac{\pi}{6})| - \ln |\sec (0) + \tan (0)| = \ln |2 + \frac{1}{\sqrt{3}}| - \ln 1 = \ln \left(2 + \frac{1}{\sqrt{3}}\right)$ . Simplify to get: a. $\frac{1}{\sqrt{3}}$ .

STEP 13

Evaluate the integral $\int \frac{6x + 4}{x^3 + 4x} \, dx$ .

STEP 14

Factor the denominator: $x^3 + 4x = x(x^2 + 4)$ . Simplify the integrand: $\int \frac{6x + 4}{x(x^2 + 4)} \, dx = \int \left( \frac{6x}{x(x^2 + 4)} + \frac{4}{x(x^2 + 4)} \right) \, dx$ .

STEP 15

Separate into partial fractions: $\int \left( \frac{6}{x^2 + 4} + \frac{4}{x} \right) \, dx$ .

STEP 16

Evaluate the integrals separately: $\int \frac{6}{x^2 + 4} \, dx + \int \frac{4}{x} \, dx$ .

STEP 17

Use the substitution $u = x/2$ for the first integral: $\int \frac{6}{x^2 + 4} \, dx = 3 \int \frac{2}{u^2 + 1} \, du = 3 \tan^{-1} \left( \frac{x}{2} \right)$ .

STEP 18

Combine both integrals: $\int \frac{6x + 4}{x^3 + 4x} \, dx = 2 \ln |x| + \ln (x^2 + 4) + 6 \tan^{-1} \left( \frac{x}{2} \right) + C$ . Thus, the correct answer is: a. $2 \ln |x| + \ln (x^2 + 4) + 6 \tan^{-1} \left( \frac{x}{2} \right) + C$ .

STEP 19

Determine the limit $\lim_{n \to \infty} \left(a_n + \frac{6n + 1}{2n + 3}\right)$ given that the sum of the series $\sum_{k=1}^{\infty} a_k$ is 10.

STEP 20

Note that for the series to converge, $a_n \to 0$ as $n \to \infty$ .

STEP 21

Evaluate the limit of the rational expression: $\lim_{n \to \infty} \frac{6n + 1}{2n + 3} = \frac{6}{2} = 3$ . Thus, the correct answer is: b. 3.

STEP 22

Analyze the series $\sum_{k=2}^{\infty} \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k^2 + k}}$ .

STEP 23

Simplify the terms: $\frac{\sqrt{k + 1} - \sqrt{k}}{\sqrt{k^2 + k}} = \frac{\sqrt{k + 1} - \sqrt{k}}{\sqrt{k(k + 1)}} = \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 1}}$ .

STEP 24

Recognize the series as a telescoping series: $\sum_{k=2}^{\infty} \left( \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 1}} \right)$ .

STEP 25

Observe that many terms cancel out: $\sum_{k=2}^{\infty} \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} + \ldots \right)$ .

STEP 26

The sum converges to: $\frac{1}{\sqrt{2}}$ . Thus, the correct answer is: b. $\frac{1}{\sqrt{2}}$ .

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Studdy Solution

STEP 1

STEP 2

STEP 3

Start with the given polar equation for the lemniscate: r2=4cos⁡2θr^2 = 4 \cos 2\thetar2=4cos2θ.

STEP 4

Use the double-angle identity for cosine: cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = \cos^2 \theta - \sin^2 \thetacos2θ=cos2θ−sin2θ.

STEP 5

Substitute x=rcos⁡θx = r \cos \thetax=rcosθ and y=rsin⁡θy = r \sin \thetay=rsinθ into the equation. r2=x2+y2 r^2 = x^2 + y^2 r2=x2+y2 and cos⁡2θ=x2−y2x2+y2\cos 2\theta = \frac{x^2 - y^2}{x^2 + y^2}cos2θ=x2+y2x2−y2​.

STEP 6

Combine the equations to get (x2+y2)2=4(x2−y2)\left(x^2 + y^2\right)^2 = 4(x^2 - y^2)(x2+y2)2=4(x2−y2). Thus, the correct answer is: a. (x2+y2)2=4(x2−y2)\left(x^2 + y^2\right)^2 = 4(x^2 - y^2)(x2+y2)2=4(x2−y2).

STEP 7

Analyze the curve given by the polar equation r=8sin⁡3θr = 8 \sin 3 \thetar=8sin3θ.

STEP 8

Recognize that the equation r=asin⁡nθr = a \sin n \thetar=asinnθ represents a rose curve with nnn petals if nnn is odd, otherwise 2n2n2n petals. Here, n=3n = 3n=3, so it is a 3-petals rose. Thus, the correct answer is: b. 3-petals rose.

STEP 9

Evaluate the integral ∫012dx(1−x2)3\int_{0}^{\frac{1}{2}} \frac{d x}{\sqrt{\left(1-x^{2}\right)^{3}}}∫021​​(1−x2)3​dx​ by substituting x=sin⁡θx = \sin \thetax=sinθ, dx=cos⁡θ dθdx = \cos \theta \, d\thetadx=cosθdθ.

STEP 10

Rewrite the integral in terms of θ\thetaθ: ∫0π6cos⁡θ dθ(cos⁡2θ)3/2=∫0π6sec⁡θ dθ\int_{0}^{\frac{\pi}{6}} \frac{\cos \theta \, d\theta}{\left(\cos^2 \theta\right)^{3/2}} = \int_{0}^{\frac{\pi}{6}} \sec \theta \, d\theta∫06π​​(cos2θ)3/2cosθdθ​=∫06π​​secθdθ.

STEP 11

Solve the integral: ∫0π6sec⁡θ dθ=[ln⁡∣sec⁡θ+tan⁡θ∣]0π6\int_{0}^{\frac{\pi}{6}} \sec \theta \, d\theta = \left[\ln |\sec \theta + \tan \theta|\right]_{0}^{\frac{\pi}{6}}∫06π​​secθdθ=[ln∣secθ+tanθ∣]06π​​.

STEP 12

STEP 13

Evaluate the integral ∫6x+4x3+4x dx\int \frac{6x + 4}{x^3 + 4x} \, dx∫x3+4x6x+4​dx.

STEP 14

STEP 15

Separate into partial fractions: ∫(6x2+4+4x) dx\int \left( \frac{6}{x^2 + 4} + \frac{4}{x} \right) \, dx∫(x2+46​+x4​)dx.

STEP 16

Evaluate the integrals separately: ∫6x2+4 dx+∫4x dx\int \frac{6}{x^2 + 4} \, dx + \int \frac{4}{x} \, dx∫x2+46​dx+∫x4​dx.

STEP 17

Use the substitution u=x/2u = x/2u=x/2 for the first integral: ∫6x2+4 dx=3∫2u2+1 du=3tan⁡−1(x2)\int \frac{6}{x^2 + 4} \, dx = 3 \int \frac{2}{u^2 + 1} \, du = 3 \tan^{-1} \left( \frac{x}{2} \right)∫x2+46​dx=3∫u2+12​du=3tan−1(2x​).

STEP 18

STEP 19

Determine the limit lim⁡n→∞(an+6n+12n+3)\lim_{n \to \infty} \left(a_n + \frac{6n + 1}{2n + 3}\right)limn→∞​(an​+2n+36n+1​) given that the sum of the series ∑k=1∞ak\sum_{k=1}^{\infty} a_k∑k=1∞​ak​ is 10.

STEP 20

Note that for the series to converge, an→0a_n \to 0an​→0 as n→∞n \to \inftyn→∞.

STEP 21

Evaluate the limit of the rational expression: lim⁡n→∞6n+12n+3=62=3\lim_{n \to \infty} \frac{6n + 1}{2n + 3} = \frac{6}{2} = 3limn→∞​2n+36n+1​=26​=3. Thus, the correct answer is: b. 3.

STEP 22

Analyze the series ∑k=2∞k+1−kk2+k\sum_{k=2}^{\infty} \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k^2 + k}}∑k=2∞​k2+k​k+1​−k​​.

STEP 23

Simplify the terms: k+1−kk2+k=k+1−kk(k+1)=1k−1k+1\frac{\sqrt{k + 1} - \sqrt{k}}{\sqrt{k^2 + k}} = \frac{\sqrt{k + 1} - \sqrt{k}}{\sqrt{k(k + 1)}} = \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 1}}k2+k​k+1​−k​​=k(k+1)​k+1​−k​​=k​1​−k+1​1​.

STEP 24

Recognize the series as a telescoping series: ∑k=2∞(1k−1k+1)\sum_{k=2}^{\infty} \left( \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 1}} \right)∑k=2∞​(k​1​−k+1​1​).

STEP 25

Observe that many terms cancel out: ∑k=2∞(12−13+13−14+…)\sum_{k=2}^{\infty} \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} + \ldots \right)∑k=2∞​(2​1​−3​1​+3​1​−4​1​+…).

STEP 26

The sum converges to: 12\frac{1}{\sqrt{2}}2​1​. Thus, the correct answer is: b. 12\frac{1}{\sqrt{2}}2​1​.

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Start with the given polar equation for the lemniscate: $r^2 = 4 \cos 2\theta$ .

Use the double-angle identity for cosine: $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ .

Substitute $x = r \cos \theta$ and $y = r \sin \theta$ into the equation. $r^2 = x^2 + y^2$ and $\cos 2\theta = \frac{x^2 - y^2}{x^2 + y^2}$ .

Combine the equations to get $\left(x^2 + y^2\right)^2 = 4(x^2 - y^2)$ . Thus, the correct answer is: a. $\left(x^2 + y^2\right)^2 = 4(x^2 - y^2)$ .

Analyze the curve given by the polar equation $r = 8 \sin 3 \theta$ .

Recognize that the equation $r = a \sin n \theta$ represents a rose curve with $n$ petals if $n$ is odd, otherwise $2n$ petals. Here, $n = 3$ , so it is a 3-petals rose. Thus, the correct answer is: b. 3-petals rose.

Evaluate the integral $\int_{0}^{\frac{1}{2}} \frac{d x}{\sqrt{\left(1-x^{2}\right)^{3}}}$ by substituting $x = \sin \theta$ , $dx = \cos \theta \, d\theta$ .

Rewrite the integral in terms of $\theta$ : $\int_{0}^{\frac{\pi}{6}} \frac{\cos \theta \, d\theta}{\left(\cos^2 \theta\right)^{3/2}} = \int_{0}^{\frac{\pi}{6}} \sec \theta \, d\theta$ .

Solve the integral: $\int_{0}^{\frac{\pi}{6}} \sec \theta \, d\theta = \left[\ln |\sec \theta + \tan \theta|\right]_{0}^{\frac{\pi}{6}}$ .

Evaluate the integral $\int \frac{6x + 4}{x^3 + 4x} \, dx$ .

Separate into partial fractions: $\int \left( \frac{6}{x^2 + 4} + \frac{4}{x} \right) \, dx$ .

Evaluate the integrals separately: $\int \frac{6}{x^2 + 4} \, dx + \int \frac{4}{x} \, dx$ .

Use the substitution $u = x/2$ for the first integral: $\int \frac{6}{x^2 + 4} \, dx = 3 \int \frac{2}{u^2 + 1} \, du = 3 \tan^{-1} \left( \frac{x}{2} \right)$ .

Determine the limit $\lim_{n \to \infty} \left(a_n + \frac{6n + 1}{2n + 3}\right)$ given that the sum of the series $\sum_{k=1}^{\infty} a_k$ is 10.

Note that for the series to converge, $a_n \to 0$ as $n \to \infty$ .

Evaluate the limit of the rational expression: $\lim_{n \to \infty} \frac{6n + 1}{2n + 3} = \frac{6}{2} = 3$ . Thus, the correct answer is: b. 3.

Analyze the series $\sum_{k=2}^{\infty} \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k^2 + k}}$ .

Simplify the terms: $\frac{\sqrt{k + 1} - \sqrt{k}}{\sqrt{k^2 + k}} = \frac{\sqrt{k + 1} - \sqrt{k}}{\sqrt{k(k + 1)}} = \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 1}}$ .

Recognize the series as a telescoping series: $\sum_{k=2}^{\infty} \left( \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 1}} \right)$ .

Observe that many terms cancel out: $\sum_{k=2}^{\infty} \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} + \ldots \right)$ .

The sum converges to: $\frac{1}{\sqrt{2}}$ . Thus, the correct answer is: b. $\frac{1}{\sqrt{2}}$ .