Calculus help?

Question

3. a right circular cone is inscribed in a sphere of radius 5. determine teh dimensions of hte cone that has the maximum possible volume

Answer 1

Let the sphere be centered at the origin, and let θ be the angle between the apex and the intersection of the base of the cone and the sphere.
r = 5sin θ
h = 5(1 - cosθ)
V = (1/3)πr^2h
V = (1/3)π(5sin θ)^2(5)(1 - cosθ)
V = (125π/3)(sin^2θ)(1 - cosθ)
V = (125π/3)(sin^2θ - sin^2θcosθ)
dV/dθ = (125π/3)( 2sinθcosθ + sin^3θ - 2sinθcos^2θ ) = 0 for min or max V
2cosθ + sin^2θ - 2cos^2θ = 0
2cosθ + 1 - 3cos^2θ = 0
3cos^2θ - 2cosθ - 1 = 0
(3cosθ + 1)(cosθ - 1) = 0
cosθ = 1, -1/3
θ = 0,π, 0.60817π
If θ = 0 or π, V = 0

using 0.60817π
V = (125π/3)(sin^2(0.60817π))(1 - cos(0.60817π))
V = (125π/3)(8/9)(1 + 1/3)
V = (125π/3)(8/9)(4/3)
V = (125π/3)(32/27)
V = 155.14

h = 20/3 ≈ 6.6667
r = 5√(8/9) ≈ 4.7140

Answer 2

The altitude of the cone h passes through the center of the sphere. Let the radius of the base of the cone be a. The cone's volume is then V = k*π*a^2*h. A right triangle is formed by the radius a, and the portion of the altitude from sphere center to cone base. The hypotenuse of that triangle is the sphere's radius r, and the leg of that triangle that lies on the cones altitude is h - r. Therefore the radius of the base of the cone a = √[r^2 - (h - r)^2]. Put this into the formula for the volume of the cone to get

V(h) = k*π*h*[r^2 - (h - r)^2]

Multiply this out to get

V(h) = k*π*h*(r^2 - h^2 - r^2 + 2*r*h)
V(h) = k*π*h*(2*r*h - h^2)
V(h) = k*π*(2*r*h^2 - h^3)

Now take the derivative and set it to zero:

d V(h) / dh = k*π(4*r*h - 3*h^2) = 0

4*r - 3*h = 0

h = (4/3)*r

This is the cone's altitude. The base of the cone is obtained from
the formula derived above, a = √[r^2 - (h - r)^2]

a = √{r^2 - [(4/3)*r - r]^2} = √{r^2 - [(1/3)*r]^2} = √[r^2 - (1/9)*r^2]

a = r*√[8/9] = (2/3)*r*√2

Answer 3

This is a tricky one. First you must understand exactly what is meant by the cone fitting inside the sphere. The circle at the base of the cone will be on the surface of the sphere. The tip of the cone will be touching the opposite side.

Now we'll have to choose a radius for the base of the cone. Let's call it r.

Draw a diagram of the setup from the side, so that you see a triangle superimposed on a circle. Now calculate the distance of the base of the cone from the centre of the circle as a function of r.

Calculate the height of the cone as a function of r.

Now calculate the volume of the cone as a function of r.

Finally, use calculus to maximise the volume.

Answer 4

Let

R = radius sphere = 5
r = radius of cone
h = height of cone
V = volume of cone

We have

V = ⅓πr²h

The height of the cone starts from the top, goes thru the center and down an additional distance. We will call that additional distance y.

h = R + y

Now we can draw a triangle

R = hypotenuse
y = one leg
r = other leg

And we have

r² + y² = R²
r² = R² - y²
y² = R² - r²

Substituting

V = ⅓πr²h = ⅓π(R² - y²)(R + y) = ⅓π(R³ + R²y - Ry² - y³)

Take the derivative to find the critical points.

dV/dy = ⅓π(R² - 2Ry - 3y²) = 0
R² - 2Ry - 3y² = 0
(R - 3y)(R + y) = 0
y = R/3,-R
y = R/3 the negative solution is rejected

h = R + y = R + R/3 = 4R/3

r² = R² - y² = R² - (R/3)² = R² - R²/9 = 8R²/9
r = (2√2/3)R

V = ⅓πr²h = ⅓π(8R²/9)(4R/3) = 32πR³/81

To summarize

h = 4R/3
r = (2√2/3)R
V = 32πR³/81

For R = 5

h = 20/3 ≈ 6.667
r = (10√2/3) ≈ 4.714
V = 4000π/81 ≈ 155.140

Answer 5

i might say shape is a factor of engineering this skill that a solid carry close on math is mandatory, pre-cal is is like algebra to the subsequent point with the Pi chart and understanding approximately radians and perspective measures (which seems important) yet Calculus is the study of derivatives, (the slope of a line at a definite factor) sort of pointless no rely what field of workd your going to, in case you inquire from me.

Answer 6

The angle made by any pt. on the circumference of the cone base and its centre at the centre of the sphere is constant. Let the angle be 'A' . So,
radius of cone (r) = 5*sinA;
ht. of cone (h) = 5*(1+cosA);
Volume of cone (V)= (1/3)*pi*(r^2)*h
= 0.333*pi*(5^3)*((sinA)^2)*(1+cosA)
=130.9*(1-(cosA)^2)*(1+cosA)
=130.9*(1+cosA-(cosA)^2-(cosA)^3)
diff. w.r.t. A
dV/dA=130.9*(1-2*cosA-3*(cosA)^2)*(-sinA)
So, extremum is when dV/dA=0
i.e. when
3*(cosA)^2 +2*(cosA)-1=0
Solving,
cosA=1/3=0.33333
So sinA = 0.94281
So the dimensions of the cone are;
radius=5*sinA=4.714;
ht. = 5*(1+cosA) =6.667;