# calculus

posted by on .

They're actaully 2 question :D

1.find the rate of change of the distance between the orgin and a moving point on the graph of y=x^2+1 if dx/dt=2 centimeters per second when x=1.

2.A triangular trough is 12 feet long and 3 feet across the top. it ends are isosceles triangles with altitude of 3 feet.
a)if water is being pumped into the trough at 2 cubic feet per minute how fast is the level rising when h is 1 ft deep?
b)if the water is rising at a rate of 3/8 inch per minute when h=2, determine the rate of which the water is being pumped into the trough.

• calculus - ,

y = x^2 + 1
dy/dt = 2x dx/dt
when x=1 and dx/dt = 2
dy/dt = 2(1)(2) = 4
Also, when x = 1, y = 2

Distance = (x^2 + y^2)^(1/2)
d(distance)/dt = (1/2)(x^2 + y^2)^(-1/2) (2x dx/dt + 2y dy/dt)

= (1/2)(1/√2) (2(2) + 2(2)(4))
= 10/√2 = 5√2

• #2 calculus - ,

Let the height of the water be h ft
let the width of the water level be 2x ft
2x/h = 3/3
h = 2x or x = h/2

V = area of triangle x 12
= (1/2)(2x)(h)(12)
= 12xh
= 12(h/2)(h) = 6 h^2

dV/dt = 12h dh/dt

a) when h = 1 and dV/dt = 2
2 = 12(1)(dh/dt)
dh/dt = 1/6 ft/min

b) You do it.