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 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

To solve the first question, we can use the concept of related rates. The rate at which the water level is rising in the trough is dependent on the rate at which water is being pumped into it. We are given that the water is being pumped at a rate of 2.4 ft3/min.

To find the rate at which the water level is rising, we need to find the derivative of the water level with respect to time (dH/dt), where H represents the depth of the water.

Let's use a similar triangle property to relate the depth of the water to the dimensions of the trough. The smaller triangle formed by the water has a height of H ft and a base of 2.6 ft. The larger triangle formed by the trough has a height of 2.1 ft and a base of 2.6 ft. Since the triangles are similar, we can set up the following proportion:

H/2.6 = 2.1/2.6

Simplifying the equation, we get:

H = (2.1/2.6) * 2.6

H = 2.1 ft

Now, let's differentiate this equation with respect to time (t):

dH/dt = d/dt (2.1 ft)

Since 2.1 ft is a constant, its derivative with respect to time will be zero. Therefore, dH/dt is equal to zero.

Hence, the water level in the trough is not changing. The rate at which the water level is rising is zero ft/min.

For the second question regarding the baseball diamond, we are given the player's speed (26 ft/s) and we need to find the rate at which the distance from home plate is changing when the player is 38 ft from second base.

Let's denote the distance from home plate as D and the distance from second base as S. We want to find dD/dt, the rate at which D is changing with respect to time.

The relationship between D and S is given by the Pythagorean theorem:

D^2 = S^2 + 90^2

Differentiating both sides of the equation with respect to time (t), we get:

2D * dD/dt = 2S * dS/dt

We are given that dS/dt = 26 ft/s and S = 38 ft. We need to find dD/dt when S = 38 ft.

Plugging in the values into the equation, we have:

2D * dD/dt = 2 * 38 * 26

2D * dD/dt = 1976

Simplifying the equation, we get:

D * dD/dt = 988

Now, we need to find D. Using the Pythagorean theorem equation:

D^2 = S^2 + 90^2

D^2 = 38^2 + 90^2

D^2 = 1444 + 8100

D^2 = 9544

D ≈ 97.68 ft

Plugging this value into the equation, we have:

97.68 * dD/dt = 988

Solving for dD/dt, we get:

dD/dt ≈ 10.11 ft/s

Hence, the rate at which the distance from home plate is changing when the player is 38 ft from second base is approximately 10.11 ft/s.