heureka

How to convert polar to cartesian and vice versa ?

I. polar to cartesian : $$\boxed{\begin{array}{rcl}
x=r*\cos{(\alpha)} \\
y=r*\sin{(\alpha)}
\end{array}
}$$

II. cartesian  to polar : $$r:
\begin{array}{rcl}
x^2+y^2&=&r^2\cos^2{(\alpha)}+r^2\sin^2{(\alpha)} \\
x^2+y^2&=& r^2(
\underbrace{
\sin^2{(\alpha)}+\cos^2{(\alpha)}
}_{=1} ) \\
x^2+y^2&=&r^2 \\
\end{array}
\boxed{r=\sqrt{x^2+y^2}}$$

$$\alpha:
\begin{array}{rcl}
\frac{y}{x}&=& \frac { r\sin{(\alpha)} } { r\cos{(\alpha)} }\\
\frac{y}{x}&=& \tan{(\alpha)}
\end{array}
\boxed{
\alpha=\tan^{-1}\left(
\frac{y}{x}
\right)
}$$

$$\begin{array}{rcll}
y>0 &and& x>0 & \quad \alpha \\
y>0 &and& x<0 & \quad \alpha +180\ensurement{^{\circ}}\\
y<0 &and& x<0 & \quad \alpha +180\ensurement{^{\circ}}\\
y<0 &and& x>0 & \quad \alpha +360\ensurement{^{\circ}}\\
\hline
y=0 &and& x>0 & \quad \alpha = 0\ensurement{^{\circ}}\\
y>0 &and& x=0 & \quad \alpha = 90\ensurement{^{\circ}}\\
y=0 &and& x<0 & \quad \alpha = 180\ensurement{^{\circ}}\\
y<0 &and& x=0 & \quad \alpha = 270\ensurement{^{\circ}}
\end{array}$$

  y = 0   and  x = 0   $$\alpha$$  undefined!

show that Ax + By = C and Ax + By = D are parallel lines explain why the restrictions on a,b,c and d are necessary.

(1) Ax+By=C

(2) Ax+By=D

y-intercept(x=0): $$(1)
\begin{array}{rcl}
\quad A*0 + By &=& C \\
By&=&C\\
y_{(x=0)}&=&\frac{C}{B}
\end{array}
\qquad
(2)
\begin{array}{rcl}
\quad A*0 + By &=& D \\
By&=&D\\
y_{(x=0)}&=&\frac{D}{B}
\end{array}$$

x-intercept (y=0):  $$(1)
\begin{array}{rcl}
\quad Ax + B*0 &=& C \\
Ax&=&C\\
x_{(y=0)}&=&\frac{C}{A}
\end{array}
\qquad
(2)
\begin{array}{rcl}
\quad Ax + B*0 &=& D \\
Ax&=&D\\
x_{(y=0)}&=&\frac{D}{A}
\end{array}$$

$$slope = -\frac{y_{(x=0)}}{x_{(y=0)}}$$

$$\text{(1) slope} = -\dfrac{
\frac{C}{B}
}{\frac{C}{A}} = -\frac{C}{C}*\frac{A}{B} = -\frac{A}{B}$$

$$\text{(2) slope} = -\dfrac{
\frac{D}{B}
}{\frac{D}{A}} = -\frac{D}{D}*\frac{A}{B} = -\frac{A}{B}$$

slope (1) = slope (2) = $$-\frac{A}{B}$$ ,  line 1 and line 2 are parallel 

Funktion f(x) gleich 3 hoch x  - Die Ableitung (Tangentenanstiege)

$$y=3^x \qquad y' \ ? \\
y=3^x \quad | \quad \ln{()}\\
\ln{(y)}=\ln{(3^x)}\\
\ln{(y)}= x\ln{(3)}\quad | \quad e^{()}\\
e^{\ln{(y)}}=y =e^{(x\ln{(3)})} \quad | \quad y=e^x \Rightarrow y'=y=e^x \\
y' = e^{(x\ln{(3)})}}
* \underbrace{\ln{(3)} }_{\text{innere Ableitung}}
\quad | \quad \text{Kettenregel\ !} \\
y'=y*\ln{(3)} = 3^x* \ln{(3)}$$

what is the minimum for this equation, C(x) 0.5x^2+23x-216+2600/x

Minimum c'(x) = 0

$$c'(x) = 2 * 0.5 x +23 -1*(2600)x^{-2}=x +23 -\frac{2600}{x^{2}} = 0\\
c'(x) = 0:\\
x +23 -\frac{2600}{x^{2}} = 0\\
\frac{2600}{x^{2}} = x+23\\
2600 = x^2(x+23)\\
2600 = x^3+23x^2\\
x^3+23x^2-2600=0\\
x= 9.01217 ( \mbox{ and 2 imaginary solutions} )$$

c(9.01217) = 320.38829

Minimum: (9.01217, 320.38829)

Find the Lcm of 32,84,180 ?

Least Common Multiple, you need the prime factorization:

$$\mbox{prime factorization} \\\\
\begin{array}{rccccc}
32 =& 2^{\textcolor[rgb]{1,0,0}{5}}\ *& 3^0 \ *& 5^0 \ *& 7^0\\
84 =& 2^2 \ *& 3^1 \ *& 5^0 \ *& 7^{\textcolor[rgb]{1,0,0}{1}}\\
180 =& 2^2 \ *& 3^{\textcolor[rgb]{1,0,0}{2}} \ *& 5^{\textcolor[rgb]{1,0,0}{1}} \ *& 7^0\\
\hline
\mbox{in each case the biggest exponent}
& 2^{\textcolor[rgb]{1,0,0}{5}}
\ *& 3^{\textcolor[rgb]{1,0,0}{2}}
\ *& 5^{\textcolor[rgb]{1,0,0}{1}}
\ *& 7^{\textcolor[rgb]{1,0,0}{1}} &= 10080 \\
\end{array}$$

LCM of 32, 84, 180 is 10080

10080 = 315 * 32

10080 = 120 * 84

10080 = 56 * 180

Given a Partial side length and two angles find the height of a given figure

The height of Mount Rushmore = H   ?

The faces on Mount Rushmore are 60 feet tall: h = 60 feet

Head at a 48 degree angle of elevation: $$\alpha = 48 \ensurement{^{\circ}}$$

His chin at a 44.76 degree angle of elevation: $$\beta = 44.76 \ensurement{^{\circ}}$$

The distance to the mountain = d

Solution:

$$\tan{(\beta)}=\frac{H-h}{d} \qquad d=\frac{H-h}{\tan{(\beta)}} \quad (1)$$

$$\tan{(\alpha)}=\frac{H}{d} \qquad d = \frac{H}{\tan{(\alpha)}} \quad (2)$$

(1) = (2):

$$\frac{H-h}{\tan{(\beta)}} =\frac{H}{\tan{(\alpha)}}$$

$$\frac {\tan{(\beta)}}{\tan{(\alpha)}} =\frac {H-h} {H} = 1- \frac{h}{H}$$

$$\frac{h}{H} = 1 - \frac {\tan{(\beta)}} {\tan{(\alpha)}}
=\frac {{\tan{(\alpha)}} -{\tan{(\beta)}} } {\tan{(\alpha)}} \quad
| \quad \updownarrow$$

$$\frac {H} {h} =\frac {\tan{(\alpha)}} {{\tan{(\alpha)}} -{\tan{(\beta)}} }$$

$$\boxed{
H = h \left(
\frac {\tan{(\alpha)}} {{\tan{(\alpha)}} -{\tan{(\beta)}} }
\right)
}$$

$$H = 60\left(
\frac {\tan{(48\ensurement{^{\circ}})}} {{\tan{(48\ensurement{^{\circ}})}} -{\tan{(44.76\ensurement{^{\circ}})}} }
\right)$$

H =  60 * 9.33639328817

H = 560.183597290 feet

The height of Mount Rushmore is 560.18 feet

Find an equation of the tangent line to the circle x2+y2=9 at the point (2, √5)

The radius r of the circle is 3, because $$x^2+y^2=r^2$$ and $$r^2 = 9$$

The distance of the point of the co-ordinate origin amounts   $$\sqrt{2^2+\sqrt{5}^2}=\sqrt{4+5}=\sqrt{9}=3 = r$$

so the point lies on the circle: The equation is: $$y=-\dfrac{x_p}{y_p}\left( x-x_p \right) +y_p$$ or $$y=-\dfrac{x_p}{y_p}x +\dfrac{r^2}{y_p}$$

so  $$x_p = 2 \quad y_p=\sqrt{5}\\
y=-\dfrac{2}{\sqrt{5}}x +\dfrac{r^2}{\sqrt{5}}$$finally  $$y=-\dfrac{2}{\sqrt{5}}x +\dfrac{9}{\sqrt{5}}$$

How do I solve for n in: 0.25 = 0.5^n

$$\begin{array}{rcl}
0.25 & = & 0.5^n\\\\
\dfrac{1}{4} &=& \left(\dfrac{1}{2} \right)^n\\\\
\dfrac{1}{2^2} &=& \dfrac{1}{2^n} \\\\
2^2 &=& 2^n \\
2&=& n
\end{array}$$

$$\boxed{n= 2}$$ 

Burgbrunnen

abwärts:  $$s=\frac{g}{2}t_1^2 \quad | \quad g=9.81\ \frac{m}{s^2}$$ 

aufwärts: $$s=ct_2\quad | \quad c=331,5\ \frac{m}{s}$$  

gleichsetzen: $$s=\frac{g}{2}t_1^2 = ct_2\quad | \quad t_1+t_2=t \qquad t=6,52s$$

t1 ersetzen durch t2: $$\frac{g}{2}(t-t_2)^2 = ct_2 \quad | \quad t_1 = t - t_2$$

Konstante k definieren: $$(t-t_2)^2=2*k*t_2 \quad | \quad k = \frac{c}{g}[s]=\frac{331,5
[\frac{m}{s}]
}
{
9,81
[\frac{m}{s^2}]
} = 33,7920489297s$$

nach t2 auflösen: $$\begin{array}{rcl}
t^2-2tt_2+t_2^2 &=& 2kt_2\\
t_2^2-2t_2(t+k)+t^2 &=& 0\\
t_2 &=& t+k\pm\sqrt{(t+k)^2-t^2} \quad | \quad t_1=t-t_2\\
t-t_2=t_1 &=& -k\mp \sqrt{(t+k)^2-t^2} \quad | \quad t_1 \;muss \;\ge 0 \;sein! \;Vorzeichen \ + \\
t_1 &=& -k+ \sqrt{(t+k)^2-t^2} \\
t_1 &=& -k+ \sqrt{\not{t^2}+2tk+k^2-\not{t^2}}
\end{array}$$

$$\textcolor[rgb]{1,0,0}{
\boxed{
\textcolor[rgb]{0,0,0}{
t_1 = -k+ \sqrt{k(k+2t)}
}
}
}$$

Lösung: $$t_1=5,98924\ s$$

$$t_2=t-t_1=0,53076\ s$$

$$s=ct_2=175,9472\ m$$

Der Brunnen ist 175,9472 m tief.

how do rationalize and simplify 2x/sqrt(8x)

$$\dfrac{2x}{ \sqrt{8x} } = \sqrt{ \dfrac{4x^2}{ 8x } } = \sqrt{ \dfrac{x}{2} } = \frac{1}{2}\sqrt{ 2x }$$