CALCULUS RELATED RATES HELP!?
c^2 = a^2 + b^2 - ab cosC c is the distance we're looking for and a and b are the lengths of the arms 2 c c' = 2 a a' + 2 b b' + ab sinC C' - (ab' + a'b) cosC we need to find the rate at which the angle between the hands is changing, C' ; we have everything else. the minute hand is moving 2 pi rad/h and the hour hand is moving at pi/6 rad/h 2 pi - pi/6 = C' = 11/6 pi all of the terms with a' or b' drop out 'cause they're zero 2 (4 cm) c' = (4 cm)(8 cm) sin(pi/6) (11/6 pi rad/h) I think you can solve for c'; it'll be in cm/h
Calculus help! (related rates)?
let x = dist from base of wall to foot of ladder y = height of ladder x^2 + y^2 = 25^2 then x dx/dt + y dy/dt = 0 2A = xy 2dA/dt = x dy/dt + y dx/dt if x = 24, then y = 7 while dx/dt = 7 thus 24(7) + 7 dy/dt = 0 dy/dt = -24 2dA/dt = (24)(-24) + 7(7) = -527 ft^2/s dA/dt = -527/2 ft^2/s ยง Edit: yeah... area of triangle... half *base*height...
Calculus Help, Related Rates??
Write an equation that relates the pile's height to time (at 5ft/min, this should be really easy). Next, write an equation that relates the volume to the height (volume of a cone, and you're given that h=2*r). Combine them to get the equation relating volume to time. Take the derivative. To do this using a pure related rates method, you don't bother with the first equation: you already have its derivative as 5ft/min (dh/dt). Take the derivative of the second (dv/dh). Multiply to get dv/dt. Finally, plug in 10 for h and solve for dv/dt.
Related rate-calculus~ HELP?
1.Animal psychologists have determined experimentally that the weight W and the surface area S of a typical horse are related by the empirical equation S=0.1W^(2/3). How fast is the surface area of a horse increasing at a time when the horse weights 350Kg and is gaining weight at the rate of 200kg/yr? 2. The revenueR a company receives is a function of the weekly sales x. also, the sales level x is a function of the weekly advertising expenditures A, and A in turn is a varying function of t, time. a) Rate of change of revenue with respect to advertising expenditures b)time rate of change of advertising expenditures c)Marginal revenue d)Rate of change of sales with respect to advertising expenditures. Write a type of chain rule that expresses the time rate of change of revenue, dR/dt in term of three of the derivatives described above. Help me! Thanks:)
Calculus help Related Rates Problems!!?!?!?
1 If A is the altitude of the plane above the antenna, and H(t) is the horizontal distance from the plane to the antenna, then the distance from the plane to the antenna is: s(t) = sqrt(H(t)^2 + A^2) For any functions f and g, if h(x) = f(g(x)) then we can use the chain rule: let u = g(x), then dh/dx = (df/du)(du/dx) = f' g' Since s = (H(t)^2 + A^2)^(1/2), assuming A is constant s'(t) = (1/2)(H(t)^2 + A^2)^(-1/2) d((H(t)^2 + A^2))/dt = (1/2)(H(t)^2 + A^2)^(-1/2)(2)(H(t))H'(t) = H(t) H'(t) / sqrt(s(t)) We know s(t) and s'(t) at the time of interest T, and we want to find H'(T). That means we need to find H(T) first since s(t) = (H(t)^2 + A^2)^(1/2) we have s(t)^2 = H(t)^2 + A^2 s(t)^2 - A^2 = H(t)^2 or H(t) = sqrt(s^2 - A^2) so we have the single equation with a single unknown: s'(T) = H'(T) sqrt(s(T)^2 - A^2)) / sqrt(s(T)) so you can solve for H'(T) #2 has the same flavor: you are given dr/dt and are being asked about dV/dt at a point specified by r = R V = (pi)(r^2)(L), where L is the length of the cylinder (constant) thus dV/dt = (dV/dr) (dr/dt) = (2)(pi)(rL) dr/dt #3 is messier but the principle is the same. You are given dV/dt and are interested in the depth y of the water which is related to the volume by: V= (pi/3)(y^2)(3R-y) Differentiating both sides gives: dV/dt = (pi/3)[6Ry - 3y^2] dy/dt so dy/dt = dV/dt (1 / [6Ry - 3y^2] You are given dV/dt, R, and y at the time of interest so you can determine dy/dt at the time of interest. the radius r of the water surface is related to the depth y by: r^2 + (R - y)^2 = R^2 r^2 + R^2 - 2Ry + y^2 = R^2 r = sqrt(2Ry - y^2) so you can compute dr/dt in terms of R, y, and dy/dt for part b and then evaluate it for y = 8 in part c.
Help on related rates questions? (calculus w/ apps)?
The area is A = (pi)(r^2). The rate of change of the area is dA/dt. From the chain rule, dA/dt = (dA/dr)(dr/dt). Now dA/dr = 2(pi)r, and dr/dt = 3 (given). So dA/dt = 2(pi)r*3 = 6 pi r. Finally, we need to use the fact that at t = 10, r will be 30. Plug that in: dA/dt = 180*pi.
Calculus Help: Related Rate, I think.?
Shampoo drips from a crack in the side of a plastic bottle at a rate modeled by: Y(t) = t/(1+t^(3/2))^1/2, where Y(t) is in ounces per minute. If there are 32 ounces in the bottle at t = 0, how many ounces are left in the bottle after 5 minutes?
Related Rates Calculus: Offshore?
Let d be the distance between them. Then, d^2 = 90^2 + (30t)^2 = 8100 + 900t^2 Differentiating with respect to time gives 2dd' = 1800t d' = 1800t / 2d We are given that d=150 feet. We know from the Pythagorean theorem that 150^2 = 90^2 + x^2 (which is the horizontal distance the shark travels). Solving for x gives 120 feet. It would take the shark 4 seconds to go this far. So now we have all we need to compute d'. d' = 1800(4) / 2(150) = 24 feet per second.
CALCULUS HELP: Related rates.. A painting hangs on the wall with the bottom...?
A painting, which is 3m tall, hangs on a wall with the bottom of the painting 3m above the floor. A man, whose eyes are 2m above the floor, looks at the painting while backing away at 0.5 m/sec. Find the rate (in radians/sec) at which the angle subtended by the painting at the man's eye is decreasing when the man is 3m from the wall. Please help me out with this question. Will award 10 points for best answer (answers only with steps)