Calculus II — Integrals

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The polar coordinates of a point are $\left(1.2, \frac{2\pi }{5}\right )$ . Give its -coordinate in the rectangular coordinate system (nearest hundredth).

Explanation

Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 1.2, \theta = \frac{2\pi }{5}$ :

$x = r \cos \theta = 1.2 \cdot \cos \frac{2\pi }{5} \approx 1.2 \cdot 0.3090 \approx 0.37$

The polar coordinates of a point are $\left(1.2, \frac{2\pi }{5}\right )$ . Give its -coordinate in the rectangular coordinate system (nearest hundredth).

Explanation

Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 1.2, \theta = \frac{2\pi }{5}$ :

$x = r \cos \theta = 1.2 \cdot \cos \frac{2\pi }{5} \approx 1.2 \cdot 0.3090 \approx 0.37$

The polar coordinates of a point are $\left(1.2, \frac{2\pi }{5}\right )$ . Give its -coordinate in the rectangular coordinate system (nearest hundredth).

Explanation

Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 1.2, \theta = \frac{2\pi }{5}$ :

$x = r \cos \theta = 1.2 \cdot \cos \frac{2\pi }{5} \approx 1.2 \cdot 0.3090 \approx 0.37$

The polar coordinates of a point are $\left(1.2, \frac{2\pi }{5}\right )$ . Give its -coordinate in the rectangular coordinate system (nearest hundredth).

Explanation

Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 1.2, \theta = \frac{2\pi }{5}$ :

$x = r \cos \theta = 1.2 \cdot \cos \frac{2\pi }{5} \approx 1.2 \cdot 0.3090 \approx 0.37$

Give the $x^{2}$ term of the Maclaurin series of the function $f \left ( x\right ) = \frac{1}{x-2}$

$-\frac{1}{8} x^{2}$

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

$-\frac{1}{12} x^{2}$

$-\frac{1}{2} x^{2}$

$-\frac{1}{6} x^{2}$

Explanation

The $x^{2}$ term of the Maclaurin series of a function has coefficient $\frac{f '' (0) }{2!} = \frac{f '' (0) }{2}$

The second derivative of can be found as follows:

$f \left ( x\right ) = \frac{1}{x-2} = (x-2) ^{-1}$

$f' \left ( x\right ) = -1\cdot (x-2)^{-2} = -1(x-2)^{-2}$

$f'' \left ( x\right ) = (-2) ( -1(x-2)^{-3}) = 2(x-2)^{-3} = \frac{2}{(x-2)^{3}}$

$f'' \left ( 0 \right ) =\frac{2}{(0-2)^{3}} = -\frac{1}{4}$

The coeficient of $x^{2}$ in the Maclaurin series is therefore

$\frac{f '' (0) }{2} = \frac{-\frac{1}{4}}{2} = - \frac{1}{8}$

Give the $x^{2}$ term of the Maclaurin series of the function $f \left ( x\right ) = \frac{1}{x-2}$

$-\frac{1}{8} x^{2}$

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

$-\frac{1}{12} x^{2}$

$-\frac{1}{2} x^{2}$

$-\frac{1}{6} x^{2}$

Explanation

The $x^{2}$ term of the Maclaurin series of a function has coefficient $\frac{f '' (0) }{2!} = \frac{f '' (0) }{2}$

The second derivative of can be found as follows:

$f \left ( x\right ) = \frac{1}{x-2} = (x-2) ^{-1}$

$f' \left ( x\right ) = -1\cdot (x-2)^{-2} = -1(x-2)^{-2}$

$f'' \left ( x\right ) = (-2) ( -1(x-2)^{-3}) = 2(x-2)^{-3} = \frac{2}{(x-2)^{3}}$

$f'' \left ( 0 \right ) =\frac{2}{(0-2)^{3}} = -\frac{1}{4}$

The coeficient of $x^{2}$ in the Maclaurin series is therefore

$\frac{f '' (0) }{2} = \frac{-\frac{1}{4}}{2} = - \frac{1}{8}$

Give the term of the Taylor series expansion of the function $f (x) = \log_5 x$ about .

$- \frac{1}{ \ln 25 } (x-1)^{2}$

$\frac{1}{ \ln 25 } (x-1)^{2}$

$\ln 25\cdot (x-1)^{2}$

$\frac{\ln 5}{ 2 } (x-1)^{2}$

$\frac{2}{ \ln 5 } (x-1)^{2}$

Explanation

The $(x-1)^{2}$ term of a Taylor series expansion about is

$\frac{f ' ' (1)}{2!} (x-1)^{2}= \frac{f ' ' (1)}{2}(x-1)^{2}$ .

We can find by differentiating twice in succession:

$f (x) = \log_5 x$

$f'(x) = \frac{1}{x \ln 5}$

$f'(x) =\frac{1}{ \ln 5} \cdot x^{-1}$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1}{ \ln 5} \cdot x^{-1} \right )$

$f''(x) =\frac{1}{ \ln 5}\cdot \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{-1} \right )$

$f''(x) =\frac{1}{ \ln 5} \cdot \left ( -1 \cdot x^{-2} \right )$

$f''(x) =- \frac{1}{ \ln 5 \cdot x^{2}}$

$f''(1) =- \frac{1}{ \ln 5 \cdot 1^{2}} = - \frac{1}{ \ln 5 }$

so the term is $\frac{f''(1) }{2} ; (x-1)^{2} = - \frac{1}{2 \ln 5 } (x-1)^{2} = - \frac{1}{ \ln 25 } (x-1)^{2}$

Give the term of the Taylor series expansion of the function $f (x) = \log_5 x$ about .

$- \frac{1}{ \ln 25 } (x-1)^{2}$

$\frac{1}{ \ln 25 } (x-1)^{2}$

$\ln 25\cdot (x-1)^{2}$

$\frac{\ln 5}{ 2 } (x-1)^{2}$

$\frac{2}{ \ln 5 } (x-1)^{2}$

Explanation

The $(x-1)^{2}$ term of a Taylor series expansion about is

$\frac{f ' ' (1)}{2!} (x-1)^{2}= \frac{f ' ' (1)}{2}(x-1)^{2}$ .

We can find by differentiating twice in succession:

$f (x) = \log_5 x$

$f'(x) = \frac{1}{x \ln 5}$

$f'(x) =\frac{1}{ \ln 5} \cdot x^{-1}$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1}{ \ln 5} \cdot x^{-1} \right )$

$f''(x) =\frac{1}{ \ln 5}\cdot \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{-1} \right )$

$f''(x) =\frac{1}{ \ln 5} \cdot \left ( -1 \cdot x^{-2} \right )$

$f''(x) =- \frac{1}{ \ln 5 \cdot x^{2}}$

$f''(1) =- \frac{1}{ \ln 5 \cdot 1^{2}} = - \frac{1}{ \ln 5 }$

so the term is $\frac{f''(1) }{2} ; (x-1)^{2} = - \frac{1}{2 \ln 5 } (x-1)^{2} = - \frac{1}{ \ln 25 } (x-1)^{2}$

Give the term of the Taylor series expansion of the function $f (x) = \log_5 x$ about .

$- \frac{1}{ \ln 25 } (x-1)^{2}$

$\frac{1}{ \ln 25 } (x-1)^{2}$

$\ln 25\cdot (x-1)^{2}$

$\frac{\ln 5}{ 2 } (x-1)^{2}$

$\frac{2}{ \ln 5 } (x-1)^{2}$

Explanation

The $(x-1)^{2}$ term of a Taylor series expansion about is

$\frac{f ' ' (1)}{2!} (x-1)^{2}= \frac{f ' ' (1)}{2}(x-1)^{2}$ .

We can find by differentiating twice in succession:

$f (x) = \log_5 x$

$f'(x) = \frac{1}{x \ln 5}$

$f'(x) =\frac{1}{ \ln 5} \cdot x^{-1}$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1}{ \ln 5} \cdot x^{-1} \right )$

$f''(x) =\frac{1}{ \ln 5}\cdot \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{-1} \right )$

$f''(x) =\frac{1}{ \ln 5} \cdot \left ( -1 \cdot x^{-2} \right )$

$f''(x) =- \frac{1}{ \ln 5 \cdot x^{2}}$

$f''(1) =- \frac{1}{ \ln 5 \cdot 1^{2}} = - \frac{1}{ \ln 5 }$

so the term is $\frac{f''(1) }{2} ; (x-1)^{2} = - \frac{1}{2 \ln 5 } (x-1)^{2} = - \frac{1}{ \ln 25 } (x-1)^{2}$

Give the $x^{2}$ term of the Maclaurin series of the function $f \left ( x\right ) = \frac{1}{x-2}$

$-\frac{1}{8} x^{2}$

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

$-\frac{1}{12} x^{2}$

$-\frac{1}{2} x^{2}$

$-\frac{1}{6} x^{2}$

Explanation

The $x^{2}$ term of the Maclaurin series of a function has coefficient $\frac{f '' (0) }{2!} = \frac{f '' (0) }{2}$

The second derivative of can be found as follows:

$f \left ( x\right ) = \frac{1}{x-2} = (x-2) ^{-1}$

$f' \left ( x\right ) = -1\cdot (x-2)^{-2} = -1(x-2)^{-2}$

$f'' \left ( x\right ) = (-2) ( -1(x-2)^{-3}) = 2(x-2)^{-3} = \frac{2}{(x-2)^{3}}$

$f'' \left ( 0 \right ) =\frac{2}{(0-2)^{3}} = -\frac{1}{4}$

The coeficient of $x^{2}$ in the Maclaurin series is therefore

$\frac{f '' (0) }{2} = \frac{-\frac{1}{4}}{2} = - \frac{1}{8}$

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