MATH SOLVE

5 months ago

Q:
# A group of engineers is building a parabolic satellite dish whose shape will be formed by rotating the curve y = ax2 about the y-axis. if the dish is to have a 10-ft diameter and a maximum depth of 3 ft, find the value of a and the surface area s of the dish. (round the surface area to two decimal places.)

Accepted Solution

A:

we have that

D=10 ft-------- > r= 5 ft

deep=3 ft

y=ax²

then

3=a*(5)²-------------- > a=3/25

y=(3/25)x²------------ > for x interval [0,5]

the answer part 1) is a=(3/25)

A differential of arc length is given by the Pythagorean theorem:

(ds)^2 = (dx)^2 +(dy)^2

then

ds = √(1 +(dy/dx)^2)·dx

dy/dx = (3/25)*(2x)=(6x/25)

Remember that when you revolve an arc around some axis, you multiply this value by the radius to get the surface area of revolution.

therefore

The dish area (A) is given by the integral

A = ∫[0, 5] 2π·x·√(1 +(dy/dx)^2)·dx

where dy/dx = (6x/25)

A = 2π/25·∫[0, 5] x√(625 +36x^2)·dx

∫x√(a +bx^2)·dx = (a +bx^2)^(3/2)/(3b)

.. = 2π/25·(625 +36x^2)^(3/2)/(3*36) |[0, 5]

.. = (π/1350)*[(625 +36*25)^(3/2) -625^(3/2)]

.. = 102.2254 ft²----------------- > 102.23 ft²

the answer part 2) is 102.23 ft²

D=10 ft-------- > r= 5 ft

deep=3 ft

y=ax²

then

3=a*(5)²-------------- > a=3/25

y=(3/25)x²------------ > for x interval [0,5]

the answer part 1) is a=(3/25)

A differential of arc length is given by the Pythagorean theorem:

(ds)^2 = (dx)^2 +(dy)^2

then

ds = √(1 +(dy/dx)^2)·dx

dy/dx = (3/25)*(2x)=(6x/25)

Remember that when you revolve an arc around some axis, you multiply this value by the radius to get the surface area of revolution.

therefore

The dish area (A) is given by the integral

A = ∫[0, 5] 2π·x·√(1 +(dy/dx)^2)·dx

where dy/dx = (6x/25)

A = 2π/25·∫[0, 5] x√(625 +36x^2)·dx

∫x√(a +bx^2)·dx = (a +bx^2)^(3/2)/(3b)

.. = 2π/25·(625 +36x^2)^(3/2)/(3*36) |[0, 5]

.. = (π/1350)*[(625 +36*25)^(3/2) -625^(3/2)]

.. = 102.2254 ft²----------------- > 102.23 ft²

the answer part 2) is 102.23 ft²