Calculus 2/Physics Question?
The volume of a cone is 1/3 π r^2 h. In this problem, r = h/2, so we can write the volume as πh^3 / 12. Since h = 8, the volume is 128π/3, and the half-tank of oil has a volume of 64π/3, and a weight of 1216π. Working backward, h = ∛(12V/π), so the height of the oil is ∛256 = 4∛4 or 2^(8/3). That's about 6 1/3 feet. Define x to be the distance from the outlet to the surface of the oil, with positive x pointing downward. Before pumping begins, x = 9 - 4∛4. When pumping is finished, x = 9. Those two values will be the integration limits. Notice that x + h = 9, where h is the height of the oil surface. The area of the oil surface is πr^2 where r = h/2 and h = 9 - x, so A = π/4 (9-x)^2. The differential element of volume is π/4 (9-x)^2 dx. Work is force times distance, so put the density in there and integrate: 57 ∫ x dV = 57π/4 ∫(9x - 18x^2 + x^3) dx = 57π/4 (9/2 x^2 - 6x^3 + 1/4 x^4) The limits are from 9 - 4∛4 to 9. When you plug in those limits, you'll have your answer. [Suggestion: From looking at those limits, I think you may get a lot of cancellation if you keep the lower limit as a binomial.]
If we drop 2 objects of different weights from the same height, which one will reach the ground faster?
In the absence of air resistance,two bodies of different masses, dropped from the same height, will touch the floor at the same time and this is true for all bodies no matter what the mass is. Since they are accelerated the same and start with the same initial conditions (at rest and dropped from a height h) they will hit the floor at the same time.However, in the real world, we have things like air resistance, which is why heavy things do fall faster. For example, if you drop a piece of paper and a pen , pen will land first since the paper is slowed down more by the air. If you did the same thing without air,both would land exactly the same time.
Calculus dimensions help?
Suppose the postal requirements specify that parcels must have length plus girth at most 60 inches. Consider the problem of finding the dimensions of the square ended rectangular package of greatest volume that is mailable. Each square end is labeled with x and the remaining dimension is labeled with h. Express the length plus the girth terms of x and h. X= 35 and h= 16 is that correct?
I need help in calculus as soon as possible?
For the water tank, the radius, r = 6 ft, and the height, h = 20 ft. Given a water depth, d = 10 ft, and water's weight density, γ = 62.5 lb/ft³, the work required to pump the water over the tank's rim is: W = γ ∫ z dV where the height the water is being pumped is "z," and the differential volume, dV, is: dV = πr² dv Therefore, the integral is: W = γ ∫ πr²z dv = γπr² ∫ z dv (evaluated over the interval [10,20]) Evaluating: W = γπr²[(1/2)z²] (evaluated over the interval [10,20]) W = (62.5)π(6)²[(1/2)(20)² - (1/2)(10)²] = 1,060,288 ft-lb This problem can also be solved directly. The work required is simply equal to the total weight of water lifted an average distance, d = (10 + 20)/2 = 15 feet. Thus, the total work is: W = γπr²hd = (62.5)π(6)²(10)(15) = 1,060,288 ft-lb (This would have been a much more interesting problem if the cylindrical tank had been lying on its side.)
Homework Question: Joe walked 4 miles north, 9 miles east, 8 miles north, and then 7 miles east. If Joe now decides to walk straight back to where he started, how far must he walk?
The answer provided by User-9597828242429937573 is great, but there are couple of omissions that make its precision unacceptable for practical purposes:1. Our planet is not spherical: its height is less than width due to its rotation. We should take it into account;2. Joe is not a point, but rather a material object. We have to make calculations probably for his geometrical center or, maybe, his center of mass. It should be discussed and decided. If agreed upon geometrical center, 1/2 of Joe's height should be added to the radius;3. Since the journey is long, the result depends on the time he starts and his speed - because of thermal expansion of the Earth's surface heated by the Sun. If Joe is walking while the surface is expanding, his trip back may be longer if the temperature is still high;4. Of course, we should consider the Earth deformation due to the Moon's gravity;5. The answer has to be expressed in closest integer number of Plank lengths. Otherwise, it would be absolutely unrealistic.Other than that, it is pretty accurate and will probably satisfy Joe. :)
Can someone help with my calculus problem involving work?
Volume of a cone is (1/3)π(r^2)h We know that r = 3h/4 because when the radius is 8, the height is 12. so V = (1/3)π(3h/4)^2*h V = (9/48)πh^3 height of the extremely dense hot chocolate is 11 so V = (9/48)π(11)^3 V = (3993π)/16 Density = m/v, so to get our mass, fill in the given d and v m = Dv = (1530 kg/m^3)((3993π)/16) m = (3054645π)/8 = 1199556.2864156kg W = Fd W = mgd W = (1199556.2864156kg)(12m) W = 14394675.436987 N
Rate of Change Questions?
h(t)=-16t^2+13t+1 v(t) = h'(t) =-32t+13=0 t=13/32 when speed is 0. h(13/32) = -16(13/32)^2 +13(13/32) +1 = 3.6406 -------------------- h(t)=t^2+14t+7 h'(t)=2t+14 h'(2) = 2(2)+14 = 18 - instantaneous rate of change of h(t) with respect to t at t=2 . ------------------------------ G(x) = 6x^2+x+4 G'(x) = 12x+1 G'(b) =12b+1 = 0 b=-1/12 --------------------------------------... g(x)=2x^2+4x+1 g(3)=2(3)^2+4(3)+1 =31 g(1)=2(1)^2+4(1)+1 = 7 Average rate of change of g(x) from x=1 to x=3 = [g(3)-g(1)](3-1) =(31-7)/2 = 12 Average rate of change = 12 Instantaneous rate of change = 4x+4 Instantaneous rate of change at x=c is 4c+4 4c+4=12 4c=8 c=2 --------------------------------------... Average rate of change of F(s) from 0 to d = [F(d)-F(0)]/d F(s) =5s^2+3s+4 F(d) =5d^2+3d+4 F(0)=4 [F(d)-F(0)]/d = [(5d^2+3d+4)-4]/d = 5d+3 Instantaneous rate of change of F(s) at s=1. F'(s) = 10s+3 F'(1)=13 5d+3=13 5d=10 d=2 --------------------------------------... Tangent line to y=f(x) is: f'(x) =2x+1 If is is parallel to the x-axis, 2x+1 = 0 (slope is 0) 2x+1=0 x=-1/2
Differential calculus rate of change problems?
1) Denote h₀ = 16 ft = height of street lamp. h₁ = 6 ft = height of Johann. x(t) = distance from Johann to lamp. s(t) = distance from the tip of Johann's shadow to lamp. By similar triangles, s(t)/h₀ = (s(t)-x(t))/h₁. So h₁s(t) = h₀(s(t)-x(t)). (h₀ - h₁)s(t) = h₀x(t). s(t) = (h₀/(h₀ - h₁))x(t). So the tip of his shadow moves at the rate s'(t) = (h₀/(h₀ - h₁))x'(t) = (16/(16 - 6))·(⁻5) = ⁻8 ft/s. (Note: s'(t) < 0 because the tip of his shadow is moving toward the lamp). His shadow changes length at rate d/dt(s(t)-x(t)) = s'(t) - x'(t) = ⁻8 - ⁻5 = ⁻3 ft/sec. (Note: the rate is negative because his shadow is shortening). 2) Denote h = 50 ft = height of lamp and ball at time 0. ℓ = 30 ft = horizontal distance from lamp to ball. y(t) = height of ball = h - at², where a = 16 ft/s² . s(t) = horizontal distance from the base of the tower to the ball's shadow. By similar triangles, s(t)/h = (s(t) - ℓ)/y(t). So s(t)y(t) = h(s(t) - ℓ). (h - y(t))s(t) = ℓh. s(t) = ℓh/(h - y(t)). So s'(t) = -ℓh·(-y'(t))/(h - y(t))² = ℓh·y'(t)/(h - y(t))² = ℓh·(⁻2at)/(at²)² = ⁻2ℓh/(at³) = ⁻2·30·50/(16·0.5³) = ⁻2·30·50/2 = ⁻1500 ft/s (Note: the rate is negative because the shadow is moving toward the tower).
What is the solution to cross the bridge within 17 minutes? (See question details below)
This is a popular question asked and answered about 4-5 years back. So far i recall, however, the problem was to find out the minimum time that all reach their camp.The thought process to decide upon the optimum strategy:• Allow always the maximum permissible persons together to cross the bridge towards their camp.• Ensure that the faster/fastest available person on the camp side to bring the torch to the rescue side.• Ensure that none of the two slowest (laziest?) persons among the four of the team (in this case the 5 & the 10) cross the bridge more than once. So they can't be allowed to cross the bridge in the first trip as well as in the last trip.• Turns out that, in the first trip, the two fastest persons in the team (in this case, the 1 and the 2) cross the bridge towards their camp. And, any one of them (ie either 1 or 2) comes back and hands over the torch to 5 and 10.• Once they arrive at camp, 2 or 1 (as the case may be), brings to rescue 1 or 2 (as the case may be).Thus, the final plan stands as below:• 1 & 2 cross; 1 OR 2 remains; 2 OR 1 returns=> Time taken: 2+2=4' OR 2+1=3'.• 5 & 10 arrives at camp and remain there. 1 OR 2 returns=> Time taken: 10+1=11 OR 10+2=12'.• 1 AND 2 arrive at camp=>Time taken=2'TOTAL TIME OF THE OPERATION:4+11+2=17'OR3+12+2=17'
A stone is thrown vertically upward with a initial speed of 19.5m/s. What is the maximum height which it covers?
v = at t = 2 sec.it takes 2 seconds for the stone to slow down amd reach max height.the stone will then fall for two seconds starting from zero speed and traveling the same distanced= 1/2at^2d = .5*9.8*2^2d = 9.8*2 = 19.6m19.6m is the answer