> There is a pole b units tall.
> a units from the top there is a point.
> How far away do you stand so the angle
> from the point to the top is maximum.
>
P_0 = base of pole
P_1 = location of point on pole = P_0 + b - a
P_2 = top of pole = P_0 + b
P_3 = where you are standing
|P_3 - P_0| = c
A = angle (P_1, P_3, P_2)
|P_2 - P_3|^2 = b^2 + c^2
|P_1 - P_3|^2 = (b - a)^2 + c^2 Pythagoreus
a^2 = b^2 + c^2 + (b - a)^2 + c^2 - 2[{(b^2 + c^2)((b - a)^2 + c^2)}^(1/2)]cos A Law of Cosines
A is the ordinate
c is the abcissa
d/dc a^2 = 0 d/dc b^2 = 0 d/dc c^2 = 2c d/dc (b - a)^2 = 0
d/dc 2[{(b^2 + c^2)((b - a)^2 + c^2)}^(1/2)]cos A
= d/dc 2[{(b^2 + c^2)((b - a)^2 + c^2)}^(1/2)]cos A + 2[{(b^2 + c^2)((b - a)^2 + c^2)}^(1/2)] d/dc cos A
= 2(1/2)[{(b^2 + c^2)((b - a)^2 + c^2)}^(-1/2)]{2c((b-a)^2 + c^2) + (b^2 + c^2)2c}cos A - 2[{(b^2 + c^2)((b - a)^2 + c^2)}^(1/2)]sin A A'
0 = 0 + 2c + 0 + 2c - [{(b^2 + c^2)((b - a)^2 + c^2)}^(-1/2)]{2c((b-a)^2 + c^2) + (b^2 + c^2)2c}cos A + 2[{(b^2 + c^2)((b - a)^2 + c^2)}^(1/2)]sin A A'
To maximize A, set A'=0
[{(b^2 + c^2)((b - a)^2 + c^2)}^(-1/2)]{2c((b-a)^2 + c^2) + (b^2 + c^2)2c}cos A=4c
[{(b^2 + c^2)((b - a)^2 + c^2)}^(-1/2)]{(b-a)^2 + c^2 + b^2 + c^2}cos A=2
cos A = (b^2 + c^2 + (b - a)^2 + c^2 - a^2)/[2[{(b^2 + c^2)((b - a)^2 + c^2)}^(1/2)]]
[{(b^2 + c^2)((b - a)^2 + c^2)}^(-1/2)]{(b-a)^2 + c^2 + b^2 + c^2}(b^2 + c^2
+ (b - a)^2 + c^2 - a^2)/[2[{(b^2 + c^2)((b - a)^2 + c^2)}^(1/2)]]=2
(2c^2 +(b - a)^2 + b^2)(2c^2 + (b - a)^2+b^2 - a^2)=4
(2c^2 +(b - a)^2 + b^2)^2 - a^2 (2c^2 +(b - a)^2 + b^2) - 4 = 0
(2c^2 +(b - a)^2 + b^2) = a^2 +/-{a^4 + 16}^(1/2) Quadradic formula
c = [{(1/2)[a^2 +/-{a^4 + 16}^(1/2) - (b - a)^2 - b^2]}^(1/2)] Solution
But is the sign on {a^4 + 16}^(1/2) positive or negative? One is the maximum
a^2 = b^2 + c^2 + (b - a)^2 + c^2 - 2[{(b^2 + c^2)((b - a)^2 +c^2)}^(1/2)]cos A
Solving the above for cos A,
cos A = (2c^2 - a^2 + b^2 + (b - a)^2)/[2{(b^2 + c^2)((b - a)^2 + c^2)}^(1/2)]
Substituting c in the above,
cos A =
(2[{(1/2)[a^2 +/-{a^4 + 16}^(1/2) - (b - a)^2 - b^2]}^(1/2)]^2 - a^2 + b^2 + (b - a)^2)/
[
[2{(b^2 + [{(1/2)[a^2 +/-{a^4 + 16}^(1/2) - (b - a)^2 - b^2]}^(1/2)]^2)((b - a)^2 + [{(1/2)[a^2 +/-{a^4 + 16}^(1/2) - (b - a)^2 - b^2]}^(1/2)]^2)}^(1/2)]
]
Let
a = 1
b = 3
cos A =
(2[{(1/2)[1 +/-{1 + 16}^(1/2) - (3 - 1)^2 - 9]}^(1/2)]^2 - 1 + 9 + (3 - 1)^2)/
[
[2{(9 + [{(1/2)[1 +/-{1 + 16}^(1/2) - (3 - 1)^2 - 9]}^(1/2)]^2)((3 - 1)^2 + [{(1/2)[1 +/-{1 + 16}^(1/2) - (3 - 1)^2 - 9]}^(1/2)]^2)}^(1/2)]
]
cos A =
(+/-17^(1/2))/
[
[2{( (1/2)[+/-17^(1/2) - 3])( (1/2)[ +/-17^(1/2) - 8])}^(1/2)]
]
cos A =
(+/-17^(1/2))/ numerator
[
[{( [+/-17^(1/2) - 3])( [ +/-17^(1/2) - 8])}^(1/2)] denominator
]
Since 17^(1/2) = 4.123105626, the sign on +/-{a^4 + 16}^(1/2) must be negative.
If it is positive the denominator is a complex number. In this particualar example,
when a=1 and b=3, Angle A = 2.030511927 radians. This isn't acceptable since
angle A must be acute. A < pi/2. That is,
if
c = [{(1/2)[a^2 -{a^4 + 16}^(1/2) - (b - a)^2 - b^2]}^(1/2)]
Multiply both sides by (-1)^2. Consequently, the distance c from the pole
resulting in the maximum angle A is,
***********************************************
c = {(1/2)[{a^4 + 16}^(1/2) + (b - a)^2 + b^2 - a^2]}^(1/2)
***********************************************
Another Method:
Apply the second derivative test and set
A'' < 0 to find whether +/-{a^4 + 16}^(1/2) is
+{a^4 + 16}^(1/2) or -{a^4 + 16}^(1/2)
That is, find
d/dc{ 4c - [{(b^2 + c^2)((b - a)^2 + c^2)}^(-1/2)]{2c((b-a)^2 + c^2) + (b^2 + c^2)2c}cos A + 2[{(b^2 + c^2)((b - a)^2 + c^2)}^(1/2)]sin A A'}=0 Equation i.
Assigning various values in Equation i.,
f_1 = cos A
f_2 = sin A A'
f_3 = 4c
f_4 = [{(b^2 + c^2)((b - a)^2 + c^2)}^(-1/2)]
f_5 = {2c((b-a)^2 + c^2) + (b^2 + c^2)2c} Equations ii.
Substituting Equations ii. into Equation i.,
d/dc{ f_3 - f_4 f_5 f_1 + 2 f_4 f_2}=0
Taking derivatives of Equations ii.,
f_1' = - sin A A' Implicit Differentiation
f_2' = cos A (A')^2 + sin A A'' Product Rule
f_3' = 4
f_4' = (-1/2)[{(b^2 + c^2)((b - a)^2 + c^2)}^(-3/2)][2c((b - a)^2 + c^2)+(b^2 + c^2)2c]
f_5' = 2((b-a)^2 + c^2)+2c(2c) + 2c(2c) + 2(b^2 + c^2) Equations iii.
Differentiating equation iii., then
d/dc{ f_3 - f_4 f_5 f_1 + 2 f_4 f_2}
= f_3' - f_4' f_5 f_1 - f_4 f_5' f_1 - f_4 f_5 f_1' + 2 f_4' f_2 + 2 f_4 f_2' =0 Equation iv.
Solve Equation iv. for sin A and square it:
f_2' = [f_3' + f_4' f_5 f_1 + f_4 f_5' f_1 + f_4 f_5 f_1' - 2 f_4' f_2]/[2f_4]
(sin A)^2 = {[f_3' + f_4' f_5 f_1 + f_4 f_5' f_1 + f_4 f_5 f_1' - 2 f_4' f_2]/[2A''f_4]}^2
1 - (cos A)^2 ={[f_3' + f_4' f_5 f_1 + f_4 f_5' f_1 + f_4 f_5 f_1' - 2 f_4' f_2]/[2A''f_4]}^2 Equation v.
f_1' = 0 at A' =0
f_2 = 0 at A' = 0 So Equation v. becomes,
1 - (f_1)^2 ={[f_3' + f_4' f_5 f_1 + f_4 f_5' f_1 ]/[2A''f_4]}^2 Equation vi. Solving for A'',
A'' = [f_3' + f_4' f_5 f_1 + f_4 f_5' f_1 ]/[2f_4[+/-{1 - (f_1)^2}^(1/2)]] Equation vii.
Substitute Equations ii. and Equations iii into Equation vii. above, and substitute
c = [{(1/2)[a^2 +/-{a^4 + 16}^(1/2) - (b - a)^2 - b^2]}^(1/2)]
Then find whether +{a^4 + 16}^(1/2) or -{a^4 + 16}^(1/2)
makes Equation vii. Negative
In the Second Derivative Test, a maximum is concave downward making the
change in the slope negative, and a minimum is concave upward making the
change in the slope positive.
cos A = (2c^2 - a^2 + b^2 + (b - a)^2)/[2{(b^2 + c^2)((b - a)^2 + c^2)}^(1/2)]
a = 1 b = 3
Y = ACOS( (2*(u^2) +12)/(2*(((9 + u^2)*( 4 + u^2))^.5) ))
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