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Calculus 3 problem: three space, prove something defines a sphere, find equation of a plane, volume, etc?

Calculus 3 problem: three space, prove something defines a sphere, find equation of a plane, volume, etc? Topic: Vector calculus homework solutions
May 19, 2019 / By Collin
Question: I have three questions: 1) Consider the points P such that the distance from P to A(-1, 5, 3) is twice the distance from P to B(6, 2, -2). Show that the set of all such points is a sphere, and find its center and radius. 2) Find an equation of the set of all points equidistant from the points A(-1, 5, 3) and B(6, 2, -2). (I know this is a plane passing through the midpoint of line segment AB and perpendicular to vector AB, but I don't know how to get a formula for this plane) 3) Find the volume of the solid that lies inside both of the spheres: x^2 + y^2 + z^2 + 4x - 2y + 4z + 5 = 0 and x^2 + y^2 + z^2 = 4 Please be specific in your answer and show me how you got the answer. This is not homework, it is from a book however, it is three of the problems that we were not assigned, but I still want to know how to do them. Thanks
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Best Answers: Calculus 3 problem: three space, prove something defines a sphere, find equation of a plane, volume, etc?

Arden Arden | 2 days ago
1.Consider the points P such that the distance from P to A(-1, 5, 3) is twice the distance from P to B(6, 2, -2). Show that the set of all such points is a sphere, and find its center and radius. ======================================... answer Let P = ( x,y,z ) , D = Distance, then 2*D ( P to B ) = D( P to A ) 4[ ( x - 6 )² + ( y - 2 )² + ( z + 2 )² ] - [ ( x + 1 )² + ( y - 5 )² + ( z - 2 )² ] 4[ x² - 12x + 36 + y² -4y +4 + z² + 4z + 4 ] - [ x² + 2x + 1 + y² - 10y + 25 + z² - 4z + 4 ] 4x² - 48x + 144 + 4y² - 16y + 16 + 4z² + 16z + 16 - x² - 2x - 1 - y² + 10y - 25 - z² + 6z - 9 = 0 3x² - 50x + 3y² - 6y + 3z² - 22z = -141 ( and divided by 3 ) x² - (50/3)x + y² - 2y + z² - (22/3)z = -14/3 x² -(50/3)x + (50/6)² - (50/6)² + y² - 2y + (1)² - (1)² + z² + (22/3)z + (22/6)² - (22/6)² = -141/3 (simplify) ( x - (50/6) )² + ( y - 1 )² + ( z + (22/6) )² = - 141/3 + 625/9 + 1 + 121/9 = 625/9 + 121/9 + 9/9 - 423/9 ( x - (50/6) )² + ( y - 1 )² + ( z + (22/6) )² = 332/9 answer center { 50/6 , 1 , 22/6 } radius { 332/9 } ======================================... 2.Find an equation of the set of all points equidistant from the points A(-1, 5, 3) and B(6, 2, -2) ======================================... let (x,y,z) be the set of points use pythagoras thereom d² = (x-a)² + (y-b)² + (z-c)² so d² = (x+1)² + (y-5)² + (z-3)² = (x-6)² + (y-2)²+ (z+2)² expand and solve gives 14x - 6y - 10z = 9 ======================================... two spheres intersect exactly when ( x + 2 )² + ( y - 1 )² + ( z + 2 )² = 4 = x² + y² + z² two spheres intersect in the plane whose equation is: D = | αx₀ + by₀ + cz₀ + d | / √(α² + b² + c²) = 9/√36 = 3/2 where a = 4, b = -2, and c = 4, d = 9, x₀ = y₀ = z₀ = 0 Considering just one sphere, say sphere 2: ( x² + y² + z² = 4 ) let r = 2, then h = ρ - α = 3/2 and h = ½ The contribution of sphere 2 to the volume of the solid is given by V₂ = ⅓πh²( 3ρ - h ) = ⅓π(½)² ( 3(2) - ½ ) = 11π/ 24 Sphere 1 contributes the same amount as sphere 2, thus the total volume: Vtotal = V₁ + V₂ since V₁ = V₂ and the solution is 2 (11π/ 24) then we got 11π/12 ( answer )
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Arden Originally Answered: The circumference of a sphere is 12.9 meters. What is the volume of the sphere?
If you have the circ you can find the diameter. C = pi x D, therefore D = c / pi 12.9 m / 3.14 = 4.10 m so the diameter is 4.10 m. Now you can use the volume formula to find the volume. Use the volume formula to solve for r. You have V = 125 and pi = 3.14 so solve for r
Arden Originally Answered: The circumference of a sphere is 12.9 meters. What is the volume of the sphere?
"The circumference of a sphere is 12.9 meters." Circumference = 2*pi*r. So that tells you r. "What is the volume of the sphere?" V = (4/3)*pi*r^3. Plug in r. "The volume of a sphere is 125 cubic feet. What is the radius of the sphere?" V = (4/3)*pi*r^3. Solve for r.
Arden Originally Answered: The circumference of a sphere is 12.9 meters. What is the volume of the sphere?
1. The circumference is equal to 2πr. So 2πr=12.9, so r=(12.9)/2π, which is approximately 2.0531. The volume of a sphere is equal to (4/3)πr^3, so plug r into the equation. This makes the volume equal (4/3)(π)(12.9/2π)^3 = (4/3)(166.41π/8π) = (4/3) (166.41/8) = 27.735 cubic meters. 2. The equation for volume is (4/3)πr^3, which in this problem is equal to 125. So 125=(4/3)πr^3. Then isolate r, 125/(4π/3)=r^3. Next, (90/π)=r^3. Then take the cubic root of both sides, so 3√(90/π) = r. r must be positive in order for the sphere to exist, so r is then approximately 3.05 feet.

Trish Trish
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Trish Originally Answered: Calculus AP Problem intersections/areas/volume?
Question - 'a': 1) Let us eliminate 'x' from both and make a combined equation in 'y' 2)==> 2y = 4x + 2b and y^2 = 4x; ==> y^2 - 2y + 2b = 0 3) This is an quadratic equation in 'y'; to have two distinct and real solutions, its discriminant must be greater than zero. 4) ==> 4 - 8b > 0; ==> b < (1/2) Answer: For all real values of 'b' less than (1/2) the two graphs will intersect at two distinct and real points Now only it struck that I have not answered your other two questions, during earlier submission; hence I am now submitting the answers for those also. Question - 'b': 1) when b = -4, the two equations are, y = 2x - 4 and y^2 = 4x. 2) Solving these two as did in previous, the two points of intersection are (1, -2) and (4, 4) For finding the area as per question, it is better to explain with a sketch; but I don't have the feasibility for making the sketch; Hence kindly refer this site:"http://www.wolframalpha.com/input/... for sketch 3) Here note the vertex is on origin, the parabola is symmetric with respect to x-axis. 4) The area enclosed region can be computed as the region enclosed between the given line, parabola, y-axis and the lines y = -2 and y =4 5) Hence the required area is: [A = Intg.{f(y)(dy)}][lim. a to b] Integral of [{(y + 4)/2} - {y^(2)/4}] dy in the limits [-2, 4] 6) On evaluation, the area is = 6 square units. Question - 'c': 1) Here also kindly refer for sketch: "http://www.wolframalpha.com/input/?i=y+... 2) On solving like previous, the point of intersection is (0, 0) and (1, 2) 3) Volume generated by rotating f(x) over x-axis is Integral [[(pi)*{f(x)^2}*(dx)] in limits[x=a, to x=b] 4) Applying this the required volume is: V = (pi)*Integral[f(x)^2 - g(x)^2] in [0, 2} { f(x) is parabola and g(x) is line} 5) V = (pi)*Integral [(4x - 4x^2)(dx)] limits[0,2] On evaluation Volume = 2(pi)/3 cubic units If in case of any doubt you may again ask with a note on this itself and if I were to go through, let me clarify.
Trish Originally Answered: Calculus AP Problem intersections/areas/volume?
(a million) For style one you will could desire to apply the chain rule and locate the by-made from sin first yet circulate away what's in the parentheses how that's: by-made from sin is cos so... cos(x^3-5x^2+10x-8) now multiply by utilising the by-made from what's in the parentheses: (3x^2-10x+10)cos(x^3-5x^2+10x-8) and that could desire to be your answer (2) For style 2 comparable ingredient: first by-made from a ability: 5(x^2sinx)^4 then multiply however the derivative on the interior: *you will could desire to apply the product rule for that* 5(x^2sinx)^4(2xsinx+x^2cosx) and that's your answer

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