Cal 1
posted by ninzyy on .
A trough is 10 ft long and 2.6 ft across the
top. Its ends are isosceles triangles with an
altitude of 2.1 ft and vertex down.
Water is being pumped into the trough at
a rate of 2.4 ft3/min.
How fast is the water level rising when the
water is 1.39 ft deep?
Answer in units of ft/min
(Sides of baseball diamond are all 90 ft)
For the baseball diamond shown in the figure
below, suppose the player is running from first
to second at a speed of 26 ft/s.
Find the rate at which the distance from
home plate is changing when the player is 38
ft from second base.

If the water is x ft deep, its crosssection is a triangle with height x and width w. By similar triangles,
x/2.1 = w/2.6
w = 1.238x
The volume is 10*(1/2)*x*w = 5*x*1.238x = 6.19x^2
v = 6.19x^2
dv/dt = 12.38x dx/dt
2.43 = 12.38(1.39) dx/dt
dx/dt = 0.14 ft/min