Posted by **ninzyy** on Thursday, November 3, 2011 at 9:39pm.

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.

- Cal 1 -
**Steve**, Friday, November 4, 2011 at 12:29pm
If the water is x ft deep, its cross-section 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

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