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.
1) To find the rate of change of the distance between the origin and a moving point on the graph of y=x^2+1, given dx/dt=2 centimeters per second when x=1, we need to use the derivative of the distance equation.
The distance between two points (x, y) and (0, 0) is given by the formula:
d = sqrt(x^2 + y^2)
Now, differentiate both sides of the equation with respect to time (t), using the chain rule:
d/dt = d(sqrt(x^2 + y^2))/dt
Since we are given dx/dt=2 centimeters per second, we can solve for dy/dt (the rate of change of y):
dx/dt = 2 = 2x * (1/sqrt(x^2 + y^2)) * dy/dt
Next, we substitute the given values:
x = 1, dx/dt = 2
We can solve for dy/dt by rearranging the equation:
2 = 2 * (1 / sqrt(1^2 + y^2)) * dy/dt
Simplifying further:
1 = (1 / sqrt(1 + y^2)) * dy/dt
dy/dt = sqrt(1 + y^2)
Since we need to find dy/dt when x = 1, we substitute x = 1 into the equation:
dy/dt = sqrt(1 + y^2) = sqrt(1 + 1^2) = sqrt(2) = approximately 1.414 centimeters per second
Therefore, the rate of change of the distance between the origin and the point on the graph y=x^2+1 is approximately 1.414 centimeters per second.
2) a) To find how fast the water level is rising when h (the depth of the water) is 1 ft in a triangular trough with given dimensions, we can use the concept of similar triangles and related rates.
Let's denote the height of the water level as H (in feet) and the length of the base of the water level as L (in feet).
From the given information, we know that the length of the trough (L) is 12 feet and the width at the top is 3 feet. Since the triangle-shaped end sections are isosceles triangles with an altitude of 3 feet, this means the base of each triangular end section is also 3 feet.
Now, let's set up a proportion based on the similar triangles. Note that the similar triangles are formed by the cross-section of the water level and the triangular end section.
H / L = 3 / 3
Simplifying the proportion:
H / L = 1
Now, we can differentiate both sides of the equation with respect to time (t) to find the related rates:
d(H) / dt = d(L) / dt
The rate at which the water level is rising (dH/dt) is what we need to determine when the depth of the water (h) is 1 ft.
b) To find the rate at which the water is being pumped into the trough when the water level is rising at a rate of 3/8 inch per minute and the depth of the water (h) is 2 ft, we can again use related rates.
Let's denote the rate at which the water is being pumped into the trough as dV/dt (in cubic feet per minute). We need to find dV/dt when h = 2 ft.
The volume of the trough is given by the formula for the volume of a triangular prism:
V = (1/2) * L * W * h
We know the length (L) is 12 feet, the width (W) is 3 feet, and the rate at which the water level is rising (dh/dt) is 3/8 inch per minute.
Now, differentiate both sides of the equation with respect to time (t) to find the related rates:
dV/dt = (1/2) * (dL/dt) * (W * h) + (1/2) * L * (dW/dt) * h + (1/2) * L * W * (dh/dt)
We are given that dh/dt = 3/8 inch per minute when h = 2 ft. Note that we need to convert 2 ft to inches for consistency.
1 ft = 12 inches
2 ft = 24 inches
Now, substitute the given values:
dV/dt = (1/2) * (dL/dt) * (3 * 24) + (1/2) * 12 * (dW/dt) * 24 + (1/2) * 12 * 3 * (3/8)
Simplifying and substituting:
dV/dt = 36 * (dL/dt) + 72 * (dW/dt) + 9/2
Therefore, to find the rate at which the water is being pumped into the trough, we would need additional information on the rates of change of the dimensions (dL/dt and dW/dt).