Water is flowing into a trough at a rate of 100 cubic centimeters per second. The trough has a length of 3 meters and cross sections in the form of a trapezoid, whose height is 50 cm, whose lower base is 25 cm and whose upper base is 1 meter. At what rate is the water rising when the depth of the water is 25 cm?
when the water has depth y cm, the width of the surface has width
25+(75/50)y cm
So, at depth y, the area of the trapezoidal cross-section is
1/2 (25 + (75/50)y)(y) = 3/4 y^2 + 25/2 y cm^2
So, the volume is
v = 9/4 y^2 + 75/2 y cm^3
So, we want dy/dt when y=25, and we know that dv/dt=100:
9/4 (2y dy/dt) + 75/2 dy/dt = 100
9/4 (50 + 75/2) dy/dt = 100
dy/dt = 32/63 cm/s
An unusual answer; better check my math.
I tried a slightly different approach
I looked at the trapezoid, and drew two verticals upwards from the base,
creating a rectangle of 25 by 50 cm and two right-angled triangles ,
each of height 50 and base 75/2
I drew a water line, of height h and width of r in each of the triangles
by ratios: r/h = (75/2) / 50 = 3/4
r = (3/4)h
V = 300( 25h + 2(1/2)rh)
= 300(25h + rh)
= 300(25h + 3/4)h^2)
dV/dt = 300( 25dh/dt + (3/2) h dh/dt)
= 300(dh/dt)(25 + (3/2)h)
when h = 25 , and dV/dt = 100
100 = 300 (dh/dt) (25 + 37.5)
dh/dt = 100/(300(62.5)) = 100/18750 = 2/375 cm/s or appr .00533 cm/s
As usual, my arithmetic should be checked also
Well, my most obvious mistake was using 3 as the length, not 300 (cm).
To find the rate at which the water is rising, we need to use the concept of related rates. Related rates involve finding the rate at which one quantity is changing with respect to another quantity that is also changing. In this case, we want to find the rate at which the water level is changing (rising), given the rate at which water is flowing into the trough.
To do this, we can apply the formula for the volume of a trapezoidal prism:
Volume = (1/2) * height * (upper base + lower base) * length
Differentiating both sides of the equation with respect to time, we get:
dV/dt = (1/2) * dh/dt * (upper base + lower base) * length
where dV/dt represents the rate at which the volume is changing with respect to time, and dh/dt represents the rate at which the depth is changing with respect to time.
Now, we know that the water is flowing into the trough at a rate of 100 cubic centimeters per second (cm^3/s):
dV/dt = 100 cm^3/s
Given that the length of the trough is 3 meters (300 cm) and the dimensions of the trapezoid (height, upper base, lower base) are constant, we can substitute these values into the equation:
100 = (1/2) * dh/dt * (1 meter + 25 cm) * 300 cm
Now, we need to find the value of dh/dt when the depth of the water is 25 cm. Let's solve for dh/dt:
100 = (1/2) * dh/dt * (1 meter + 25 cm) * 300 cm
Dividing both sides of the equation by [(1/2) * (1 meter + 25 cm) * 300 cm], we get:
dh/dt = 100 / [(1/2) * (1 meter + 25 cm) * 300 cm]
Simplifying the equation further, we have:
dh/dt = 100 / [(1/2) * (100 cm + 25 cm)]
dh/dt = 100 / [(1/2) * 125 cm]
dh/dt = 100 / (1/2) * 125 cm
dh/dt = 100 / (1/2) * 125 cm
dh/dt = 100 / (1/2) * 125 cm
dh/dt = 100 / 62.5 cm
dh/dt = 1.6 cm/s
Therefore, when the depth of the water is 25 cm, the water level is rising at a rate of 1.6 cm/s.