BeMath-1-x-tan-1-x-dx-2-Find-the-distance-of-the-point-3-3-1-from-the-plane-with-equation-r-i-j-i-j-k-0-also-find-the-point-on-the-plane-that-

Question Number 108099 by bemath last updated on 14/Aug/20

\frac{★ B e M a t h ⊚}{⊓}

(1) \int x \tan^{- 1} (x) d x ?

(2) F i n d t h e d i s t a n c e o f t h e p o i n t

(3, 3, 1) f r o m t h e p l a n e Π w i t h e q u a t i o n

(\vec{r} - \vec{i} - \vec{j}) ∙ (\vec{i} - \vec{j} + \vec{k}) = 0, a l s o

f i n d t h e p o i n t o n t h e p l a n e t h a t i s

n e a r e s t t o (3, 3, 1) .

Commented by bemath last updated on 14/Aug/20

t h a n k y o u b o t h

Answered by Dwaipayan Shikari last updated on 14/Aug/20

\int {x t a n}^{- 1} x

\frac{x^{2}}{2} {t a n}^{- 1} x - \frac{1}{2} \int \frac{x^{2}}{1 + x^{2}}

\frac{x^{2}}{2} {t a n}^{- 1} x - \frac{x}{2} + \frac{1}{2} {t a n}^{- 1} x + C

Answered by mathmax by abdo last updated on 14/Aug/20

1) A = \int x \arctan (x) dx by parts

A = \frac{x^{2}}{2} arctanx - \int \frac{x^{2}}{2} \frac{dx}{1 + x^{2}}

= \frac{x^{2}}{2} \arctan (x) - \frac{1}{2} \int \frac{1 + x^{2} - 1}{1 + x^{2}} dx

= \frac{x^{2}}{2} \arctan (x) - \frac{x}{2} + \frac{1}{2} \arctan (x) + C

= \frac{1}{2} (x^{2} + 1) arctanx - \frac{x}{2} + C

Answered by 1549442205PVT last updated on 15/Aug/20

\vec{i} = (1, 0, 0), \vec{j} = (0, 1, 0), \vec{k} = (0, 0, 1)

\Rightarrow \vec{i} - \vec{j} + \vec{k} = (1, - 1, 1), \vec{r} = (x, y, z)

\Rightarrow \vec{r} - \vec{i} - \vec{j} = (x - 1, y - 1, z)

(\vec{r} - \vec{i} - \vec{j}) ∙ (\vec{i} - \vec{j} + \vec{k})

= (x - 1, y - 1, z) ∙ (1, - 1, 1) = 0 \Leftrightarrow x - 1 - y + 1 + z = 0

\Leftrightarrow x - y + z = 0

d [(3, 3, 1), Π] = \frac{∣ 3 - 3 + 1 ∣}{\sqrt{1^{2} + {(- 1)}^{2} + 1^{2}}} = \frac{1}{\sqrt{3}}

ii) Let M (x, y, y - x) \in Π be any point of

the plane Π which has the equation :

x - y + z = 0 and A (3, 3, 1) . Then

d (A, M) = \sqrt{{(x - 3)}^{2} + {(y - 3)}^{2} + {(y - x - 1)}^{2}}

we need to find smallest of the expression

P = {(x - 3)}^{2} + {(y - 3)}^{2} + {(y - x - 1)}^{2}

= {2 x}^{2} + {2 y}^{2} - 4 x - 8 y - 2 xy + 19

= {2 x}^{2} - (2 y + 4) x + {2 y}^{2} - 8 y + 19

= 2 [{(x - \frac{y + 2}{2})}^{2} + {2 y}^{2} - 8 y + 19 - \frac{{(y + 2)}^{2}}{2}

= 2 [{(x - \frac{y + 2}{2})}^{2} + \frac{{3 y}^{2} - 20 y + 34}{2}

= 2 [{(x - \frac{y + 2}{2})}^{2} + \frac{3 {(y - \frac{10}{3})}^{2} + \frac{2}{3}}{2}

2 [{(x - \frac{y + 2}{2})}^{2} + \frac{3 {(y - 2)}^{2}}{2} + \frac{1}{3} ⩾ \frac{1}{3}

\Rightarrow d (A, M) ⩾ \frac{1}{\sqrt{3}}

The equality ocurrs if and only if

{\begin{cases} y - \frac{10}{3} = 0 \\ x - \frac{y + 2}{2} = 0 \end{cases} \Leftrightarrow {\begin{cases} x = 8 / 3 \\ y = 10 / 3 \end{cases} \Rightarrow z = 2 / 3

Thus the nearest distance between the

point A (3, 3, 1) and Π be \frac{1}{\sqrt{3}} and the point

M lie on the plane Π which is nearest to

A having coordinate be M (\frac{8}{3}, \frac{10}{3}, \frac{2}{3})

other way

Let H (x, y, y - x) \in Π . Then A (3, 3, 1) is nearest

to the plane Π if and only if \vec{AH} ⊥ Π

\Leftrightarrow \vec{AH} / / \vec{n_{Π}} = (1, - 1, 1), since \vec{AH} = (x - 3, y - 3, y - x - 1),

so \vec{AH} / / \vec{n_{Π}} \Leftrightarrow \frac{x - 3}{1} = \frac{y - 3}{- 1} = \frac{y - x - 1}{1}

\Leftrightarrow {\begin{cases} x + y = 6 \\ 2 x - y = 2 \end{cases} \Leftrightarrow {\begin{cases} x = 8 / 3 \\ y = 10 / 3 \\ z = 2 / 3 \end{cases} \Rightarrow H (\frac{8}{3}, \frac{10}{3}, \frac{2}{3})

The distance between A and the plane

Π equal to d (A, H) = \sqrt{{(3 - \frac{8}{3})}^{2} + {(3 - \frac{10}{3})}^{2} + {(1 - \frac{2}{3})}^{2}}

= \sqrt{3 {(\frac{1}{3})}^{2}} = \frac{1}{\sqrt{3}}