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The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.
First we will draw a right angle triangle ΔABC, where ∠B = 90°
Lets, assume the height(AB) = h and base(BC)= x
If the angle of elevation, ∠ACB = α, then
tan(α) =
Taking inverse trigonometric function, α = tan⁻¹ () .............(1)
As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :
Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,
that means h = 40 feet and = 10 feet/second.
C is the observer's position which is 50 feet away from the point B, so x = 50 feet.
= 0.12 (Rounding up to two decimal places)
So, the rate of change of the angle of elevation is 0.12 radians/second.
The formula for the radius r in terms of x . and for the maximum areas is x=2/+4
Given that,
y forms a circle of radius r
y=2r
r=y/2
(2-y)- forms Square Side x
(2-y) = 4x
x=(2-y)/4
Now Sum of Area's=Area of Square +Area of Circle
Sum = r² + x²
Substitute the r and x values in above equation,
A(y)= y²/4+(y-2)²/ 16
To maximize Area A(y)
A'(y)= 0
2y/4 + 2(y-2)/16 =0
y/2 + (y-2)/8 =0
y = 2/
+4
Y max will be max, x to be maximum.
for maximum sum of areas,
x=2/+4
Hence,The formula for the radius r in terms of x . and for the maximum areas is x=2/+4
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