Since the beacon makes one revolution (360°) in 10 sec.

At the point of observation, let 0 be the angle of deviation of the beacon of light from OA.
Let AB x km,when \(\angle\)AOB = 0
\(\frac { d\theta }{ dt } =\frac { 2\pi }{ 10 } \)
\(=\frac { \pi }{ 5 } \) rad /sec.
Let AB = x
Then, tan θ \( =\frac { x }{ 5 } \)
x = 5 tan θ
We know that velocity = \(\frac { dx }{ dt } \)
Differentiating (1) with respect to 't' we get,
\(\frac { dx }{ dt } =5{ sec }^{ 2 }\theta ,\frac { d\theta }{ dt } \)
\(=5({ sec }^{ 2 }{ 45 }^{ o })\left( \frac { \pi }{ 5 } \right) \)
\(=5({ \sqrt { 2 } })^{ 2 }\left( \frac { \pi }{ 5 } \right) \)
\(=5(2)\left( \frac { \pi }{ 5 } \right) \)
= 2π km/sec.

We have, \(P=\frac{234+16x}{x+3}\)
Therefore, \(\frac{dP}{dt}= - \frac{186}{(x+3)^{2}}\times \frac{dx}{dt}\).
Substituting \(x=90, \frac{dx}{dt}=15\) we get\(\frac{dP}{dt}= -\frac{186}{93^{2}}\times 15= -\frac{10}{31}\approx -0.32\) repee/ week.
That is the price is changing, in fact decreasing at the rate of Rs. 0.32 per unit.

The average rate of change in an interval [a, b] is \(\frac { f(b)-f(a) }{ b-a } \) whereas, the instantaneous rate of change at a point x is f′(x) for the given function. They are respectively, b + a and 2x.

We have, \(\frac{dy}{dx}=2x+3\). Hence at (1, 2), \((\frac{dy}{dx})=5\)

Therefore, the required equation of tangent is.
\((y-2)=5(x-1)\Rightarrow 5x-y-3=0\) 
The slope of the normal at the point (1, 2) is -\(\frac{1}{5}\).
therefore, the required equation of normal is
\((y-2)=-\frac{1}{5}(x-1)\Rightarrow x+5y-11=0\) 

Given m (x) = \(\sqrt { 3 } x\) = \(\sqrt { 3 } .{ x }^{ \frac { 1 }{ 2 } }\)
Differentiating with respect to 'x' we get,

when x = 3, \(\frac { dm }{ dx } =\frac { \sqrt { 3 } }{ 2\sqrt { 3 } } =\frac { 1 }{ 2 } \) Kg/m
when x = 27, \(\frac { dm }{ dx } =\frac { \sqrt { 3 } }{ 2\sqrt { 27 } } \)
\(=\frac{\sqrt{\not 3}}{2(3) \sqrt{\not 3}}=\frac{1}{6} \mathrm{Kg} / \mathrm{m}\)

Given s = 2t² + 3t
s(3) = 2 \(\times\) 3² + 3 (3)
= 2\(\times\)9+9
= 27 m ....(1)
s(6) = 2\(\times\) 6² + 3 (6)
= 72 + 18 = 90m ... (2)
Average velocity = \(\frac { s(6)-s(3) }{ 6-3 } \)
= \(\frac { 90-27 }{ 3 } \) = 21 m/s
(ii) Instantaneous Velocity V(t) = \(\frac { ds }{ dt } \)
Instantaneous Velocity at t = 3
= V(3) = 15 m/sec  
Instantaneous Velocity at t = 6
= V(6) = 27m/sec

Distance moved in time 't' is s = \(\frac{t^{3}}{3}-t^{2}+3\) 
Velocity at time 't ' is V = \(\frac{ds}{dt}=t^{2}-2t\)
Acceleration at time 't ' is a(t) = \(\frac{dV}{dt}=2t-2\)
Therefore, the velocity is zero when t² − 2t = 0, that is t = 0, 2. The acceleration is zero when 2t − 2 = 0 . That is at time at time t = 1

Let r be the radius of the ripple and A be the area of the ripple.
GIven \(\frac { dr }{ dt } \) = 2 cm/sec and r = 5 cm ... (1)
We know A = πr²
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } \) = π(2r).\(\frac { dr }{ dt } \)
= π(2) (5) (2) [using (1)]
\(\frac { dA }{ dt } \) = 20 πsq.cm/sec.

(i) At the maximum height, the velocity v(t) of the particle is zero.
Now, we find the velocity of the particle at time t.
\(v(t)=\frac{ds}{dt}=128-32t\)
\(v(t)=0 \Rightarrow 128-32t=0 \Rightarrow t=4.\)
After 4 seconds, the particle reaches the maximum height.
The height at t = 4 is s(4) = 128(4) - 16(4)² = 256 ft.
(ii) When the particle hits the ground then s = 0 .
s = 0 ⇒ 128t −16t² = 0
⇒ t = 0, 8 seconds.
The particle hits the ground at t = 8 seconds. The velocity when it hits the ground v(8) = –128 ft /s.