1)A helicopter takes off from a point 120m away from an abserver located on the ground, and rises vertically at a constant rate of 2.5m/s.At what rate is the elvation angle of the observer's ine of sight to te helipoter changeing ehrn the helicopter is 90m above the ground (Assume the height from the ground, of the observe's sight level is neglible)?
2.)A piece of wire of length 100cm is cut into two parts.One parts is bent into a square, while the other part is bent into a circle.The length of a side of the sqaure is x cm.Find the value of x which gives the least total area.What is the least total area?
+129849 1) The tangent of the angle at any time is given by:
tan(θ) = y / 120 differentiating this with respect to time, t, we have
sec^2(θ) dθ/dt = (1/120) dy/dt (1)
When the helicopter is 90m above the ground, θ is given by
arctan(90/120) = about .6435 rads ....so....filling in the missing pieces in (1), we have
sec^2(.6435) dθ/dt = (1/120m) (2.5m/s)
dθ/dt = (1/120m) (2.5m/s)/ sec^2(.6435) = about .01334 rads/s
2) The area of the square = x^2
The perimeter of the circle = 100 - 4x = 2*pi*r
Therefore.....the radius of the circle is [50 -2x] /pi
So....the total area is given by :
A = x^2 + pi ( [50 -2x] /pi )^2 = x^2 + (1/pi) *(50-2x)^2
Take the derivative of this and set it to 0
A' = 2x - 2* (2/pi) [ 50 - 2x] =
2x + 8[x -25] /pi = 0
x + 4[x - 25]/ pi = 0
pi*x + 4x - 100 =0
x (pi + 4) = 100
x = 100 / (pi + 4) = about 14.002 cm
The second derivative =
1 + 4/pi .....since this is always positive, the curve is concave up at all points....so x = 14.002cm is a minimum
And this is a minimum as verified by the curve shown here : https://www.desmos.com/calculator/orqtlce6ju
The minimum total area is about 350.062 cm^2
+129849 1) The tangent of the angle at any time is given by:
tan(θ) = y / 120 differentiating this with respect to time, t, we have
sec^2(θ) dθ/dt = (1/120) dy/dt (1)
When the helicopter is 90m above the ground, θ is given by
arctan(90/120) = about .6435 rads ....so....filling in the missing pieces in (1), we have
sec^2(.6435) dθ/dt = (1/120m) (2.5m/s)
dθ/dt = (1/120m) (2.5m/s)/ sec^2(.6435) = about .01334 rads/s
2) The area of the square = x^2
The perimeter of the circle = 100 - 4x = 2*pi*r
Therefore.....the radius of the circle is [50 -2x] /pi
So....the total area is given by :
A = x^2 + pi ( [50 -2x] /pi )^2 = x^2 + (1/pi) *(50-2x)^2
Take the derivative of this and set it to 0
A' = 2x - 2* (2/pi) [ 50 - 2x] =
2x + 8[x -25] /pi = 0
x + 4[x - 25]/ pi = 0
pi*x + 4x - 100 =0
x (pi + 4) = 100
x = 100 / (pi + 4) = about 14.002 cm
The second derivative =
1 + 4/pi .....since this is always positive, the curve is concave up at all points....so x = 14.002cm is a minimum
And this is a minimum as verified by the curve shown here : https://www.desmos.com/calculator/orqtlce6ju
The minimum total area is about 350.062 cm^2
