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# How to solve this RELATED RATES problem?

Topic: How to solve this RELATED RATES problem?
June 26, 2019 / By Christel
Question: Sand is being dumped into a conical pile whose height is 1/2 the radius of its base. Suppose sand is being pumped at a rate 5 cubic meters per minute. ANSWER the following QUESTIONS: (a) How fast is the height of the pile increasing when it is 9 meters height? (b) How fast is the area of the base increasing at this moment? (c) How fast is the circumference of the base increasing at this moment? (d) Will the height be increasing more slowly, more rapidly or at steady pace as time goes on?

## Best Answers: How to solve this RELATED RATES problem?

Aurora | 3 days ago
v = (π/3)r^2h dv/dt = (π/3)[r^2(dh/dt) + 2h(dr/dt)] Since the height is one-half the radius of the base, h = r/2, r = 2h, dr/dt = 2(dh/dt). a) h = 9, r = 18. 5 = (π/3)[18^2(dh/dt) + 2(9)(2)(dh/dt)] 5 = (π/3)[324(dh/dt) + 36(dh/dt)] 5 = (π/3)(360)(dh/dt) 5 = (120π)(dh/dt) dh/dt = (5/120π) = 0.01326 meters/min b) a = πr^2 da/dt = 2πr(dr/dt) = 2πr(2)(dh/dt) =2π(18)(2)(.01326) = 3 square meters per minute c) c = 2πr dc/dt = 2π(dr/dt) = 2π(2)(dh/dt) = 2π(2)(.01326) = 0.16666 meters per minute d) dv/dt = (π/3)[r^2(dh/dt) + 2h(dr/dt)] dv/dt = (π/3)[r^2(dh/dt) + 2h(2)(dh/dt)] 5 = (π/3)(dh/dt)[r^2 + 4h] dh/dt = 5 / [(π/3)(r^2 + 4h)] This shows that the height will be increasing more slowly as time goes on because the denominator of the rate of change is getting larger (r and h are increasing).
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Originally Answered: How to solve Calculus related rates problem?
No calculus needed ! Draw yourself a diagram and take it one step at a time - His tangential velocity V is radius * angular-velocity^2 1 rev per 2 mins = angulat velocity of 2 pi / 120 radians/sec The horiziontal and vertical velocitiy components when the radius is at an angle A with the horizontal are -V sinA and V cos A At 8 meters above ground, sin A = (8 - 6 ) / 6 = 1/3
Originally Answered: How to solve Calculus related rates problem?
is that this a calculus difficulty? would not look like it. First calculate the circumference of the ferris wheel which may well be 2 x pi x radius or 2 x 3.142 x 6 = 37.704m If its rotating at a million revolution consistent with 2 minutes any element on the wheel strikes at 18.852m/min or 0.3142m/sec or a million.131km/hr If a rider is 8m above the floor he remains shifting at a million.131km/hr do you are able to confirm the horizontal and vertical velocities? if so you will could use algebra.

Abbygael
Let x = radius height = x/2 V = ⅓πx²(x/2) V = (1/6)πx³ dV/dt = (1/2)πx²(dx/dt) = 5 m³/min (a) How fast is the height of the pile increasing when it is 9 meters height? h = 9m = x/2 x = 18 (1/2)πx²(dx/dt) = 5 dx/dt = 2(5)/πx² = 10/π(18²) = 0.0098 m/min dh/dt = (1/2)dx/dt = 0.0049 m/min (b) How fast is the area of the base increasing at this moment? A = πx² dA/dt = 2πx(dx/dt) = 2π(18)(0.0098) = 1.108 m²/min (c) How fast is the circumference of the base increasing at this moment? C = 2πx dC/dt = 2π(dx/dt) = 2π(0.0098) = 0.062 m/min (d) Will the height be increasing more slowly, more rapidly or at steady pace as time goes on more slowly because dh/dt has the radius squared in the denominator which increase as time goes by so that dh/dt increases more slowly.
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Stanley
V = (1/3)πr²h a) differentiate implicitly and find h' b) A=πr², A'=2πrr' c) C=2πr, C' = 2πr' d) find h'(t) and its trend as t -> ∞
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Originally Answered: Related rates rectangle problem (solving for the rates of change of the area, perimeter.)?
i) a = lw da/dt = l*dw/dt + w*dl/dt da/dt = 12*2 - 5*2 da/dt = 14 cm^2/sec ii) p = 2l + 2w dp/dt = 2*dl/dt + 2*dw/dt dp/dt = -2*2 + 2*2 dp/dt = 0 cm/sec iii) d = sqrt(l^2 + w^2) dd/dt = (1/2)(l^2 + w^2)^(-1/2)(2l*dl/dt + 2w*dw/dt) dd/dt = (1/2)(12^2 + 5^2)^(-1/2)(2*12*(-2) + 2*5*2) dd/dt = -1.077 cm/sec

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