5485 Shares

# Help with related rates problem (Calculus)? Topic: Help with related rates problem (Calculus)?
July 19, 2019 / By Cymbeline
Question: A cylinder is being filled such that the height of its contents is increasing at a rate of 5 cm/sec. What is the rate of change of the volume if r=3 and V= pi*r^2*h? this problem doesn't seem that difficult, but there seems to be some info missing... or maybe I'm not solving it correctly? Here's another one A heap of garbage in the shape of a cube is being compacted. Given that the volume decreases at a rate of 2 cubic meters per minute, find the rate of change of an edge of the cube when the volume is exactly 27 cubic meters. for the second problem I got that the edge would be changing with a rate of approximately -.01235 meters per second... but I do not know if that is correct. I think my second answer is wrong.. time to evaluate lol ## Best Answers: Help with related rates problem (Calculus)? Aubry | 1 day ago
V = pi r^2 h d/dt(V) = d/dt(pi r^2 h) dV/dt = pi r^2 dh/dt...notice the radius doesn't change with time dV/dt = pi 3^2 5 dV/dt = 45pi cm/s For the second one, let s = cube's side length. V = s^3 d/dt(V) = d/dt(s^3) dV/dt = 3s^2(ds/dt) Find s by substituting 27 for V in V = s^3 27 = s^3 s = 3 Substitute 3 for s and -2 for dV/dt and solve for ds/dt -2 = 3(3^2)ds/dt -2 = 27ds/dt ds/dt = -2/27 m/min
👍 130 | 👎 1
Did you like the answer? Help with related rates problem (Calculus)? Share with your friends Originally Answered: Calculus related rates problem?
Volume of cylinder = pi*r²*h dV/dt = pi(r²*dh/dt + 2rh*(dr/dt)) We know that dr/dt = 0 and dV/dt = 3 m^3/min 3 m^3/min = pi(r²)(dh/dt) 3 m^3/min = pi*25m²*dh/dt dh/dt = 3/(25*pi) m/min Originally Answered: Calculus related rates problem?
In calculus terms, you are told that dV/dt = 3 m^3/min, and asked to find dh/dt. dh/dt = (dh/dV)(dV/dt) V = pi r^2 h for a cylinder, so h = V/25pi dh/dV = 1/25pi so dh/dt = 3/25pi m/min Vianne
1.) There seems to be a parameter missing from this problem. If the height is given, it could be solved. 2.) Volume, V = 27 cu. m. dV/dt = - 2 cu. m./min. (Negative because it is decreasing) Edge Length = s Find ds/dt: s = ³√V s = ³√27 s = 3 V = s³ Differentiating Implicitly Over Time, dV/dt = 3s²(ds/dt) - 2 = 3(3)²(ds/dt) - 2 = 3(9)(ds/dt) - 2 = 27(ds/dt) ds/dt = - 2 / 27 ds/dt = - 0.074 The edge length is decreasing at a rate of 0.074 m., or 74 mm./min. ...
👍 50 | 👎 -2 Serrena
[a] floor section = circumference * top = 2? r * h top = 10cm + 0.1cm * t h = 10 + 0.1t A = 10? * (10 + 0.1t) = 100? + ? t dA/dt = value of replace of section with relation to time dA/dt = (100? + ? t)' = ? cm^2/sec [b] volume = component of base * top = ? r^2 * (10 + 0.1t) V = 25? (10 + 0.1t) V = 250? + 2.5? t dV/dt = 2.5? cm^3/sec desire this helps!
👍 50 | 👎 -5 Originally Answered: Calculus problem involving related rates? Most of the problem solved, find my mistake?
c = 13.748 Differentiate with respect to time, 2cc' = 2absin60° (θ)' Solve for c', c' = 12(15)sin60° (2°)(pi/180°)/c = 2 pisin60/c = pi√(3)/13.748 = .396 m/min ----------- Attention: You don't need to convert 60 deg into pi/3 rad for sin60, but you have to convert deg/min to rad/min.

If you have your own answer to the question Help with related rates problem (Calculus)?, then you can write your own version, using the form below for an extended answer.