Question-181619

Question Number 181619 by mr W last updated on 27/Nov/22

Commented by mr W last updated on 27/Nov/22

a g o a t i s c o n n e c t e d t h r o u g h a r i n g w i t h

a r o p e o f 30 m l e n g t h w h o s e e n d s a r e

f i x e d a t t w o p o i n t s w i t h a d i s t a n c e o f

20 m a s s h o w n . a g a t e a t p o i n t C o n

t h e f e n c e i s 20 m a p a r t f r o m p o i n t A .

f i n d t h e m a x i m u m a r e a w h e r e t h e

g o a t c a n g r a z e . w h a t i s t h e s m a l l e s t

d i s t a n c e t h e g o a t c a n g e t t o t h e g a t e .

Answered by Acem last updated on 27/Nov/22

Commented by Acem last updated on 27/Nov/22

F a r t h e s t d i s t a n c e i s x_{1}^{'} n o t t h e p o i n t c_{δ = 32.01 m}

F a r ._{d i s} = 34.07 m; x_{1}^{'} = 15.24 m

Commented by Acem last updated on 27/Nov/22

c ∙ A d d i t i o n a l q u e s t i o n :

A s s u m e t h a t t h e g a t e w i d e i s 6 m, a n d w e w a n t

t o m a k e t h a t g o a t c a n g e t t o m i d d l e o f i t,

{w h a t}^{'} s t h e l e n g t h o f t h e r o p e s h o u l d b e ? w i t h t h e

s a m e t w o p r e v i o u s a n c h o r p o i n t s .

T h e b e g i n i n g o f t h e g a t e i s 20 m a w a y f r o m t h e p o i n t A

Commented by mr W last updated on 27/Nov/22

t h a n k s s i r!

Commented by Acem last updated on 28/Nov/22

T h a n k y o u t o o S i r!

Answered by mr W last updated on 27/Nov/22

Commented by mr W last updated on 27/Nov/22

t h e m a x i m u m a r e a t h e g o a t c a n r e a c h

i s a n e l l i p s e w i t h s e m i a x e s :

a = 15

b = \sqrt{15^{2} - 10^{2}} = 5 \sqrt{5}

{(\frac{x}{15})}^{2} + {(\frac{y}{5 \sqrt{5}})}^{2} = 1

\Rightarrow y = 5 \sqrt{5} \sqrt{1 - {(\frac{x}{15})}^{2}}

A = 2 \times 5 \sqrt{5} \int_{- 10}^{15} \sqrt{1 - {(\frac{x}{15})}^{2}} d x

= 75 \sqrt{5} {[\sin^{- 1} \frac{x}{15} + \frac{x}{15} \sqrt{1 - {(\frac{x}{15})}^{2}}]}_{- 10}^{15}

= 75 \sqrt{5} (\frac{π}{2} + \sin^{- 1} \frac{2}{3} + \frac{2 \sqrt{5}}{9}) \approx 469.14 m^{2}

s a y P (- a \cos θ, b \sin θ)

\frac{10 - 15 \cos θ}{20 - 5 \sqrt{5} \sin θ} = \frac{5 \sqrt{5} \cos θ}{15 \sin θ}

\Rightarrow \frac{3}{\cos θ} - \frac{2 \sqrt{5}}{\sin θ} = 2

\Rightarrow θ \approx 1.1249

d_{m i n} = \sqrt{{(10 - 15 \cos θ)}^{2} + {(20 - 5 \sqrt{5} \sin θ)}^{2}}

\approx 10.29 m

Commented by mr W last updated on 27/Nov/22

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