Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 189304 by ajfour last updated on 14/Mar/23

Commented by ajfour last updated on 14/Mar/23

$F i n d l a r g e s t r a d i u s b a l l b e n e a t h$ $c o n e a n d w i r h i n c y c l i n d e r .$
	Answered by mr W last updated on 14/Mar/23

	Commented by mr W last updated on 14/Mar/23

	Commented by mr W last updated on 15/Mar/23

	$\tan β = \frac{b}{2 a}$ $m = \tan (β + θ) = \frac{\frac{b}{2 a} + \tan θ}{1 - \frac{b \tan θ}{2 a}} = \frac{b + 2 a \tan θ}{2 a - b \tan θ}$ $c e n t e r o f b a l l :$ $C [(a - r) \cos ϕ, (a - r) \sin ϕ, r]$ $v e r t e x o f c o n e :$ $B (- a, 0, 0)$ $a x i s o f c o n e :$ $\vec{B G} = (1, 0, m)$ $\vec{B C} = ((a - r) \cos ϕ + a, (a - r) \sin ϕ, r)$ $B C = \sqrt{{[(a - r) \cos ϕ + a]}^{2} + {(a - r)}^{2} \sin^{2} ϕ + r^{2}}$ $B C = \sqrt{a^{2} + {(a - r)}^{2} + r^{2} + 2 a (a - r) \cos ϕ}$ $\cos (α + θ) = \frac{\vec{B G} \cdot \vec{B C}}{∣ \vec{B G} ∣∣ \vec{B C} ∣} = \frac{(a - r) \cos ϕ + a + m r}{B C \times \sqrt{1 + m^{2}}}$ $B C \cos α \cos θ - B C \sin α \sin θ = \frac{(a - r) \cos ϕ + a + m r}{\sqrt{1 + m^{2}}}$ $\cos θ \sqrt{a^{2} + {(a - r)}^{2} + 2 a (a - r) \cos ϕ} - r \sin θ = \frac{(a - r) \cos ϕ + a + m r}{\sqrt{1 + m^{2}}}$ $l e t λ = \frac{r}{a}$ $\Rightarrow \cos θ \sqrt{1 + {(1 - λ)}^{2} + 2 (1 - λ) \cos ϕ} - λ \sin θ = \frac{(1 - λ) \cos ϕ + 1 + m λ}{\sqrt{1 + m^{2}}}$ $o r$ $[(1 + m^{2}) \cos^{2} θ - {(m + \sin θ \sqrt{1 + m^{2}} - \cos ϕ)}^{2}] λ^{2} - 2 (1 + \cos ϕ) [(1 + m^{2}) \cos^{2} θ + m + \sin θ \sqrt{1 + m^{2}} - \cos ϕ] λ + 2 (1 + \cos ϕ) (1 + m^{2}) \cos^{2} θ - {(1 + \cos ϕ)}^{2} = 0$ $\Rightarrow λ = \frac{(1 + \cos ϕ) [(1 + m^{2}) \cos^{2} θ + m + \sin θ \sqrt{1 + m^{2}} - \cos ϕ] - \sqrt{{(1 + \cos ϕ)}^{2} {[(1 + m^{2}) \cos^{2} θ + m + \sin θ \sqrt{1 + m^{2}} - \cos ϕ]}^{2} - [(1 + m^{2}) \cos^{2} θ - {(m + \sin θ \sqrt{1 + m^{2}} - \cos ϕ)}^{2}] [2 (1 + \cos ϕ) (1 + m^{2}) \cos^{2} θ - {(1 + \cos ϕ)}^{2}]}}{(1 + m^{2}) \cos^{2} θ - {(m + \sin θ \sqrt{1 + m^{2}} - \cos ϕ)}^{2}}$ $λ_{m a x} i s n o t a t ϕ = 0 ° a s f o l l o w i n g$ $e x a m p l e s h o w s .$ $e x a m p l e : a = 1, b = 1, θ = 30 °$ $\Rightarrow r_{m a x} \approx 0.3823 a t ϕ \approx 28.1826 °$
	Commented by mr W last updated on 14/Mar/23

	Commented by ajfour last updated on 23/Mar/23
	Sir i had gotten busy, so i posted my solution after a week and thank you incredible effort you have put in. I truly believe your answer should be right, but can u cast a glance and help in finding mistake in mine ..
	Commented by mr W last updated on 12/Apr/23

	$s o r r y i s a w t h i s m e s s a g e r i g h t n o w,$ $b e c a u s e i {d o n}^{'} t g e t n o t i f i c a t i o n$ $a u t o m a t i c a l l y .$ $i^{'} l l t r y t o f o l l o w y o u r s o l u t i o n .$
	Answered by ajfour last updated on 22/Mar/23

	Commented by ajfour last updated on 23/Mar/23
	$tan θ=m tan β=(b/(2a)) tan (θ+β)=q=((m+(b/(2a)))/(1−((bm)/(2a))))=((2am+b)/(2a−mb)) AJ=AD=s BJ=ms BH=r+ms DP=(r+ms)cos δ ABsin (θ+β)=r+(r+ms)sin δ AB=(s/(cos θ))=s(√(1+m^2 )) ⇒ ((ssin (θ+β))/(cos θ))=r+(r+ms)sin δ ......(i) let CP=p=a−AP AP=ABcos (θ+β) ⇒ p=a−((scos (θ+β))/(cos θ)) cos φ=((DP^2 −(a−r)^2 −p^2 )/(2p(a−r))) also cos φ=((s^2 −(a−r)^2 −a^2 )/(2a(a−r))) let (r/a)=x , (s/a)=y , (p/a)=z ⇒ (x+my)^2 cos^2 δ−(1−x)^2 −z^2 =z{y^2 −(1−x)^2 −1} ....(ii) p=a−((scos (θ+β))/(cos θ)) ⇒ z=1−((ycos (θ+β))/(cos θ)) ((ssin (θ+β))/(cos θ))=r+(r+ms)sin δ ⇒ (x+my)^2 sin^2 δ={((ysin (θ+β))/(cos θ))−x}^2 Adding to (ii) (x+my)^2 = (1−x)^2 +z^2 +z{y^2 −(1−x)^2 −1} +{((ysin (θ+β))/(cos θ))−x}^2 now z=1−((ycos (θ+β))/(cos θ)) ⇒ (x+((ysin θ)/(cos θ)))^2 = (1−x)^2 +{1−((ycos (θ+β))/(cos θ))}^2 +{1−((ycos (θ+β))/(cos θ))}{y^2 −(1−x)^2 −1} +{((ysin (θ+β))/(cos θ))−x}^2 ⇒ x^2 +((y^2 sin^2 θ)/(cos^2 θ))+((2xysin θ)/(cos θ))=(1−x)^2 +1 +((y^2 cos^2 (θ+β))/(cos^2 θ))−((2ycos (θ+β))/(cos θ)) +y^2 −1−(1−x)^2 +x^2 +((y^2 sin^2 (θ+β))/(cos^2 θ))−((2xysin (θ+β))/(cos θ)) +{1+(1−x)^2 −y^2 }((ycos (θ+β))/(cos θ)) ⇒ ((2xy)/(cos θ)){sin θ+sin (θ+β)} =2y^2 +((y(x^2 −2x−y^2 )cos (θ+β))/(cos θ)) ⇒ 2x{sin θ+sin (θ+β)} =2ycos θ+(x^2 −2x−y^2 )cos (θ+β) differentiating w.r.t. y with (dx/dy)=0 ⇒ 2cos θ=2ycos (θ+β) ⇒ y=((cos θ)/(cos (θ+β))) ⇒ 2x_m {sin θ+sin (θ+β)}cos (θ+β) =x_m (x_m −2)cos^2 (θ+β)+cos^2 θ If θ=30° tan β=(b/(2a))=(1/2) , then tan (θ+β)=((2+(√3))/(2(√3)−1)) hence 2x_m {(1/2)+((2+(√3))/(2(√5)))}(((2(√3)−1))/(2(√5))) =x_m (x_m −2)(((2(√3)−1)^2 )/(20))+(3/4) ⇒ 2x_m {((2(√3)−1)/(4(√5)))+(((2(√3)−1)^2 )/(10))} =(((2(√3)−1)^2 x_m ^2 )/(20))+(3/4) ⇒ x_m ^2 −2x_m {((√5)/(2(√3)−1))+2}+((15)/((2(√3)−1)^2 ))=0 say (2(√3)−1)x_m =q q^2 −2q((√5)+4(√3)−2)+15=0 ⇒ q ≈ 1.1371 x_m =(r_m /a)=(q/((2(√3)−1)))≈ 0.4615$
	$\tan θ = m$ $\tan β = \frac{b}{2 a}$ $\tan (θ + β) = q = \frac{m + \frac{b}{2 a}}{1 - \frac{b m}{2 a}} = \frac{2 a m + b}{2 a - m b}$ $A J = A D = s B J = m s$ $B H = r + m s$ $D P = (r + m s) \cos δ$ $A B \sin (θ + β) = r + (r + m s) \sin δ$ $A B = \frac{s}{\cos θ} = s \sqrt{1 + m^{2}}$ $\Rightarrow \frac{s \sin (θ + β)}{\cos θ} = r + (r + m s) \sin δ$ $. . . . . . (i)$ $l e t C P = p = a - A P$ $A P = A B \cos (θ + β)$ $\Rightarrow p = a - \frac{s \cos (θ + β)}{\cos θ}$ $\cos ϕ = \frac{D P^{2} - {(a - r)}^{2} - p^{2}}{2 p (a - r)}$ $a l s o$ $\cos ϕ = \frac{s^{2} - {(a - r)}^{2} - a^{2}}{2 a (a - r)}$ $l e t \frac{r}{a} = x, \frac{s}{a} = y, \frac{p}{a} = z$ $\Rightarrow {(x + m y)}^{2} \cos^{2} δ - {(1 - x)}^{2} - z^{2}$ $= z {y^{2} - {(1 - x)}^{2} - 1} . . . . (i i)$ $p = a - \frac{s \cos (θ + β)}{\cos θ}$ $\Rightarrow z = 1 - \frac{y \cos (θ + β)}{\cos θ}$ $\frac{s \sin (θ + β)}{\cos θ} = r + (r + m s) \sin δ$ $\Rightarrow$ ${(x + m y)}^{2} \sin^{2} δ = {\frac{y \sin (θ + β)}{\cos θ} - x}^{2}$ $A d d i n g t o (i i)$ ${(x + m y)}^{2} =$ ${(1 - x)}^{2} + z^{2} + z {y^{2} - {(1 - x)}^{2} - 1}$ $+ {\frac{y \sin (θ + β)}{\cos θ} - x}^{2}$ $n o w z = 1 - \frac{y \cos (θ + β)}{\cos θ}$ $\Rightarrow$ ${(x + \frac{y \sin θ}{\cos θ})}^{2} =$ ${(1 - x)}^{2} + {1 - \frac{y \cos (θ + β)}{\cos θ}}^{2}$ $+ {1 - \frac{y \cos (θ + β)}{\cos θ}} {y^{2} - {(1 - x)}^{2} - 1}$ $+ {\frac{y \sin (θ + β)}{\cos θ} - x}^{2}$ $\Rightarrow$ $x^{2} + \frac{y^{2} \sin^{2} θ}{\cos^{2} θ} + \frac{2 x y \sin θ}{\cos θ} = {(1 - x)}^{2} + 1$ $+ \frac{y^{2} \cos^{2} (θ + β)}{\cos^{2} θ} - \frac{2 y \cos (θ + β)}{\cos θ}$ $+ y^{2} - 1 - {(1 - x)}^{2}$ $+ x^{2} + \frac{y^{2} \sin^{2} (θ + β)}{\cos^{2} θ} - \frac{2 x y \sin (θ + β)}{\cos θ}$ $+ {1 + {(1 - x)}^{2} - y^{2}} \frac{y \cos (θ + β)}{\cos θ}$ $\Rightarrow$ $\frac{2 x y}{\cos θ} {\sin θ + \sin (θ + β)}$ $= 2 y^{2} + \frac{y (x^{2} - 2 x - y^{2}) \cos (θ + β)}{\cos θ}$ $\Rightarrow$ $2 x {\sin θ + \sin (θ + β)}$ $= 2 y \cos θ + (x^{2} - 2 x - y^{2}) \cos (θ + β)$ $d i f f e r e n t i a t i n g w . r . t . y w i t h$ $\frac{d x}{d y} = 0 \Rightarrow$ $2 \cos θ = 2 y \cos (θ + β)$ $\Rightarrow y = \frac{\cos θ}{\cos (θ + β)}$ $\Rightarrow$ $2 x_{m} {\sin θ + \sin (θ + β)} \cos (θ + β)$ $= x_{m} (x_{m} - 2) \cos^{2} (θ + β) + \cos^{2} θ$ $I f θ = 30 ° \tan β = \frac{b}{2 a} = \frac{1}{2}, t h e n$ $\tan (θ + β) = \frac{2 + \sqrt{3}}{2 \sqrt{3} - 1}$ $h e n c e$ $2 x_{m} {\frac{1}{2} + \frac{2 + \sqrt{3}}{2 \sqrt{5}}} \frac{(2 \sqrt{3} - 1)}{2 \sqrt{5}}$ $= x_{m} (x_{m} - 2) \frac{{(2 \sqrt{3} - 1)}^{2}}{20} + \frac{3}{4}$ $\Rightarrow 2 x_{m} {\frac{2 \sqrt{3} - 1}{4 \sqrt{5}} + \frac{{(2 \sqrt{3} - 1)}^{2}}{10}}$ $= \frac{{(2 \sqrt{3} - 1)}^{2} x_{m}^{2}}{20} + \frac{3}{4}$ $\Rightarrow$ $x_{m}^{2} - 2 x_{m} {\frac{\sqrt{5}}{2 \sqrt{3} - 1} + 2} + \frac{15}{{(2 \sqrt{3} - 1)}^{2}} = 0$ $s a y (2 \sqrt{3} - 1) x_{m} = q$ $q^{2} - 2 q (\sqrt{5} + 4 \sqrt{3} - 2) + 15 = 0$ $\Rightarrow q \approx 1.1371$ $x_{m} = \frac{r_{m}}{a} = \frac{q}{(2 \sqrt{3} - 1)} \approx 0.4615$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum