Calculus review problems

[degreeplus]

1) The line normal to the curve \(\displaystyle y = \sqrt {16 - x} \\)at the point (0,4) has slope
(A) 8
(B) 4
(C) 1/8
(D) -1/8
(E) -8

so I began by finding \(\displaystyle y' = \frac{{ - 1}}{{2\sqrt {16 - x} }}\\)

then i plugged in 0 to find (D)-1/8 but the answer is (A)8. So im wondering what is this question asking for? slope of the tangent line to \(\displaystyle \
y = \sqrt {16 - x}
\\)?

I have trouble solving this problem as well.

2) \(\displaystyle \int\limits_1^\infty {\frac{x}{{(1 + x^2 )^2 }}} dx\\) is
(A) -1/2
(B) -1/4
(C) 1/4
(D)1/2
(E) divergent

here I tried taking the limit -> \(\displaystyle \
{\lim }\limits_{b \to \infty } \int\limits_1^b {\frac{x}{{(1 + x^2 )^2 }}} dx
\\)=\(\displaystyle \
{\lim }\limits_{b \to \infty } \frac{1}{2}\int\limits_1^b {\frac{{2x}}{{(1 + x^2 )^2 }}} dx
\
\\)
$\displaystyle u = (1 + x^2 )$
du=2x dx

from here I forget how to solve. i know I need the natural log but the squared part bothers me so I'm thinking I messed up solving some place because I would get divergent but the answer is (C) 1/4.

And here I have trouble finding the third-degree Taylor polynomial for this problem

3)Let $\displaystyle f$ be the function given by $\displaystyle f(x) = ln(3-x)$. The third-degree Taylor polynomial for $\displaystyle f$ about x=2 is
(A) \(\displaystyle -(x-2) + \frac{{(x - 2)^2 }}{2}\ - \frac{{(x - 2)^3 }}{3}\\)
(B) \(\displaystyle -(x-2) - \frac{{(x - 2)^2 }}{2}\ - \frac{{(x - 2)^3 }}{3}\\)
(C) $\displaystyle (x-2) + (x - 2)^2 + (x - 2)^3$
(D) \(\displaystyle (x-2) + \frac{{(x - 2)^2 }}{2}\ + \frac{{(x - 2)^3 }}{3}\\)
(E) \(\displaystyle (x-2) - \frac{{(x - 2)^2 }}{2}\ - \frac{{(x - 2)^3 }}{3}\\)

I figured since $\displaystyle ln (1+x) = x - \frac{{(x)^2}}{2}\ + \frac{{(x)^3}}{3}\ -...$ I'm not sure if this is how you go about solving the problem but then I would center it at x=2 and multiply the series by -1. This way I would get (A) but the answer is (B).

Sorry if this seems like a list of problems but I have tried resolving these on my own with no success.
Can some one point me in the right direction? Thanks for any help

 

[skeeter]

1) the slope of a normal line is perpendicular to the tangent line ... hence the opposite reciprocal value ... -1/8 and 8.

2) after substituting, you have du/u² ... antiderivative is -1/u

(1/2) lim{b->inf} [-1/(1+b²) + 1/(1 + 1)] = 1/4

3) ln(1+x) = x - x²/2 + x³/3 - x⁴/4 + ...

ln(3 - x) = ln[1 + (2 - x)] = ln[1 + -(x-2)] = -(x - 2) - [-(x - 2)]²/2 + [-(x - 2)]³/3 + [-(x - 2)]⁴/4 + ...

what happens now when you simplify?

 

[galactus]

[quote:2jvzxyjv]1) The line normal to the curve \(\displaystyle \y = \sqrt {16 - x} \\)at the point (0,4) has slope
(A) 8
(B) 4
(C) 1/8
(D) -1/8
(E) -8

so I began by finding \(\displaystyle y' = \frac{{ - 1}}{{2\sqrt {16 - x} }}\\)

then i plugged in 0 to find (D)-1/8 but the answer is (A)8. So im wondering what is this question asking for? slope of the tangent line to \(\displaystyle \
y = \sqrt {16 - x}
\\)?[/quote:2jvzxyjv]

The normal means it is perpendicular to the curve which is tangent at (0,4). So, the slope would be the negative reciprocal of -1/8.

I have trouble solving this problem as well.

2) \(\displaystyle \int\limits_1^\infty {\frac{x}{{(1 + x^2 )^2 }}} dx\\) is
(A) -1/2
(B) -1/4
(C) 1/4
(D)1/2
(E) divergent

here I tried taking the limit -> \(\displaystyle \
{\lim }\limits_{b \to \infty } \int\limits_1^b {\frac{x}{{(1 + x^2 )^2 }}} dx
\\)=\(\displaystyle \
{\lim }\limits_{b \to \infty } \frac{1}{2}\int\limits_1^b {\frac{{2x}}{{(1 + x^2 )^2 }}} dx
\
\\)
$\displaystyle u = (1 + x^2 )$
du=2x dx

from here I forget how to solve. i know I need the natural log but the squared part bothers me so I'm thinking I messed up solving some place because I would get divergent but the answer is (C) 1/4. [/quote]

\(\displaystyle \L\\\int\frac{x}{(x^{2}+1)^{2}}dx=\frac{-1}{2(x^{2}+1)}\right]_{1}^{L}\)

\(\displaystyle \L\\\lim_{L\to\infty}\left[\frac{-1}{2(L^{2}+1)}+\frac{1}{2(1^{2}+1)}\right]=\frac{1}{2(1^{2}+1)}=\frac{1}{4}\)

And here I have trouble finding the third-degree Taylor polynomial for this problem

3)Let $\displaystyle f$ be the function given by $\displaystyle f(x) = ln(3-x)$. The third-degree Taylor polynomial for $\displaystyle f$ about x=2 is
(A) \(\displaystyle {-}(x-2) + \frac{{(x - 2)^2 }}{2}\ - \frac{{(x - 2)^3 }}{3}\\)
(B) \(\displaystyle {-}(x-2) - \frac{{(x - 2)^2 }}{2}\ - \frac{{(x - 2)^3 }}{3}\\)
(C) $\displaystyle (x-2) + (x - 2)^2 + (x - 2)^3$
(D) \(\displaystyle (x-2) + \frac{{(x - 2)^2 }}{2}\ + \frac{{(x - 2)^3 }}{3}\\)
(E) \(\displaystyle (x-2) - \frac{{(x - 2)^2 }}{2}\ - \frac{{(x - 2)^3 }}{3}\\)
I figured since $\displaystyle ln (1+x) = x - \frac{{(x)^2}}{2}\ + \frac{{(x)^3}}{3}\ -...$ I'm not sure if this is how you go about solving the problem but then I would center it at x=2 and multiply the series by -1. This way I would get (A) but the answer is (B).

Sorry if this seems like a list of problems but I have tried resolving these on my own with no success.
Can some one point me in the right direction? Thanks for any help

Follow the series formula.

\(\displaystyle \L\\P_{n}=f(2)+f'(a)(x-2)+\frac{f''(2)}{2!}(x-2)^{2}+\frac{f'''(2)}{3!}(x-2)^{3}+........\)

 

[degreeplus]

Thanks for the help!