calculus
posted by Yoona .
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.

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 
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.
Respond to this Question
Similar Questions

calculus
find the rate of change of the distance between the origin and a moving point on the graph of y=x^2+2 if ds/dt=5 centimeters per second. 
calculus
find the rate of change of the distance between the origin and a moving point on the graph of y=x^2+2 if ds/dt=5 centimeters per second. 
calculus
Find the rate of change of the distance between the origin and a moving point on the graph y = x^2 + 8 if dx/dt = 8 centimeters per second. 
calculus
6. A trough is in the shape of a triangular prism. It is 5 feet long and its vertical cross sections are isosceles triangles with base 2 feet and height 3 feet. Water is being siphoned out of the trough at the rate of 2 cubic feet … 
Calculus
find the rate of change of the distance between the orgin and a moving point on the graph of y=x^2+1 idf dx/dt=2 centimeters per second. 
Calculus
A particle is moving along the curve . As the particle passes through the point , its coordinate increases at a rate of units per second. Find the rate of change of the distance from the particle to the origin at this instant. 
Calculus
A particle is moving along the curve . As the particle passes through the point , its coordinate increases at a rate of units per second. Find the rate of change of the distance from the particle to the origin at this instant. 
4 CALCULUS study problem
1. How fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 15 cubic centimeters per second? 
Calculus Can you help me solve for dr/dt
How fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 15 cubic centimeters per second? 
math
t = time d = distance Alana walking rate: 1 meter per second Gilberto walking rate: 2 meters per second Leanne walking rate: 2.5 rating per second Alana Equation: d = t Gilberto Equation: d = 2t Leanne Equation: d = 2.5t I made a table …