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Solutions to various calculus problems from a fall 2014 calculus i class. The problems involve finding the rates of change of areas, volumes, and distances in different scenarios, such as isosceles right triangles, swimming pools, inverted conical water tanks, and hot-air balloons. The solutions use calculus concepts like derivatives and integrals.
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Math 180: Calculus I Fall 2014
TA: Brian Powers
Since the legs of the isosceles right triangle are in ratio of 1 :
2 to the hypotenuse, the rate of change of the legs will be proportional, i.e. x = h/
dh dt
Because the area A = x^2 /2, dA dt
dA dx
dx dt
= x(− 2
2 x m/s
When x = 5, dA/dt = − 2
2 m^2 /s. (b) At what rate are the length of the legs of the triangle changing? SOLUTION: The legs are decreasing at a constant rate of 2
2 m/s. (c) At what rate is the area of the triangle changing when the area is 4 m^2? SOLUTION: The area is 4 m^2 when x^2 /2 = 4 ⇒ x^2 = 8 so x = 2
2)^2 = −8 m^2 /s.
16-2 TA Discussion 16: October 23
The volume when height h ≤ 2 has been filled up is then V (h) = 20 12 (25h)h = 250h^2. For 2 < h ≤ 3, V (h) = 250(2)^2 + 20(50)(h − 2) = 1000h − 1000.
V (h) =
250 h^2 if h ≤ 2 1000 h − 1000 if 2 < h ≤ 3
After 250 minutes, the volume will be 250 m^3 , which means h = 1. We have
dV dt
dV dt
dV dh
dh dt
= 2 · 250 h dh dt
= 500 when h = 1
Which means that 1 = 500
dh dt So dhdt = 1/500 m/min, or 2mm/min (b) How long will it take to fill the pool? SOLUTION: This is easier. We just have to calculate the total volume. The cross section of the pool has area 1 · 50 + 12 · 50 · 2 = 100m^2 , so the volume is 100 · 20 = 2000 m^3 , so it takes 2000 minutes to fill the pool.
So the volume of water is V (h) =
π(. 5 h)^2 h =
π 12
h^3
Furthermore, we have
−2 =
dV dt
dV dh
dh dt
π 4
h^2
dh dt We can solve for dh/dt = − 8 /(πh^2 ), so when h = 3, dh/dt = − 8 /(9π) ft/sec.