Solid Geometry #6 - Questions and Answers

Question 1 What is a vertex of a cone? Answer There are no vertices. A vertex is a point on a 3D figure where line segments (edges) meet. A cone has no sides. The round middle is not an edge. Question 2 How many edges will a cube have if we cut 2 corners out of it? Answer A cube would gain three edges if a corner were cut off of it. Therefore, two corners would give us six additional edges in addition to the original twelve, giving us a total of 18. Question 3 The height of a cylinder is increasing at a constant rate of 10 meters per minute, and the volume is increasing at a rate of 1135 cubic meters per minute. At the instant when the height of the cylinder is 9 meters and the volume is 354 cubic meters, what is the rate of change of the radius? The volume of a cylinder can be found with the equation V = πr²h. Round answer to three decimal places Answer Given: height of a cylinder is increasing at a constant rate of 10m/min. i.e dh/dt = 10m/min (h: height of cylinder) Volume is increasing at a rate of 1135 cubic meters/min. dV/dt = 1135 m³/min At the instant: V = 354m³ h = 9 m Relation between volume (V), radius (r) and height (h) V = πr²h Differentiate above with respect to time dV/dt = d/dt(πr²h) (Take constant outside and apply product rule) dV/dt = πd/dt(r²h) dV/dt = π [r² * (dh/dt) + h d(r²)/dt] dV/dt = π[r²*(dh/dt) + h * 2r * (dr/dt)] Given: dh/dt = 10 m/min dV/dt = 1135 m³/min At the instant: V = 354 i.e [πr²h = 354 m³ h=9m πr²(9)= 354 r² = 354/(9π) r = √(354/(9π)) 1135 = π((354/(9π))10+ 2√(354/(9π))9dr/dt) 1135 = 3540/9+ 18π√(354/(9π))*dr/dt dr/dt = (1135-3540/9) / (18π√(354/(9π))) dr/dt ≈ 1.413 m/min Question 4 A Norman window of maximum area can be constructed by adjoining a semicircle to the top of an ordinary rectangular window. The dimensions of the Norman window would be such that the radius is x/2 and the total perimeter is 22 feet. Answer Let y represent the measure of the vertical dimension of the rectangular portion of the window. Let x represent the horizontal dimension of the window. x is also the diameter of the semi-circular part of the window. The perimeter of the window is then: 22 = x + 2y + (1/2)πx P_Total = (P_Rectangle - x) + P_Circle 22 = x + 2y + (1/2)πx 10 - (π/5)x = y A = xy + (π(x/2)²)/2 A_Total = A_Rectangle + A_Circle A = x(10 - (π/5)x) + (π/8)x² A = 10x - (π/5)x² + (π/8)x² Plug in y = 10 - (π/5)x for y A = 10x - (π/5)x² + (π/8)x² dA/dx = 10 - (2π/5)x + (2π/8)x Power rule d/dx (xⁿ) = nxⁿ⁻¹ 0 = 10 - (2π/5)x + (π/4)x Set dA/dx equal to 0 0 = 10 - (8πx + 5πx)/20 x + (π/5)x = 10 x(1 + π/5) = 10 x = 10 / (1 + π/5) x = 10 / ((5 + π)/5) x = 50 / (5 + π) = 40 / (4 + π) d²A/dx² = -(2π/5) + (2π/8) = (4π - 5π)/10 = -π/10 < 0 Solve for y: y = 10 - (π/5)x y = 10 - (π/5) (40 / (4 + π)) = 10 - (8π / (4 + π)) = (40 + 10π - 8π)/(4 + π) = (40 + 2π) / (4 + π) Hence x = 40 / (4 + π) is the x-coordinate of the maximum point. The y dimension and the semi-circle radius are calculated as above. Question 5 Find the volume of the solid in the first octant bounded by the coordinate planes, the plane x = 3, and the parabolic cylinder z = 4 - y². Answer ∫∫∫ f(x,y)dA R To find the value of double integral, we must found the area of integration. It is given: f(x, y) = 4 - y² The domain is restricted by the area x=3. Now, we can solve the integral: ∫[0,2] ∫[0,3] (4 - y²) dx dy = ∫[0,2] [4x - y²x] [0, 3] dy ∫[0,2] (12 - 3y²) dy = [12y - y³] [0, 2] = [12(2) - (2)³] - [0] = 24 - 8 = 16 Results: 16