(1.a) At what rate is his distance from second base decreasing when he is halfway to first base? m/s

Draw the figure.

Place the the batter somewhere between homeplate and first base .

Let x be his distance from the homeplate.

Let D be his distance from the 2nd base.

So, by Pythagorean theorem,

D = sqrt[90^2 +(90 -x)^2]

D^2 = 90^2 +(90 -x)^2

Differentiate both sides with respect to time t,

2D dD/dt = 2(90 -x)(-1) dx/dt

D dD/dt = (x -90)dx/dt -----------**

When x = 90/2 = 45m, we know dx/dt = 25m/sec, but we don't know D. So we find D.

D = sqrt[90^2 +(90 -45)^2] = 100.623 m

Hence,

(100.623)dD/dt = (45 -90)(25)

dD/dt = (-45)(25)/100.623 = -11.18 m/sec

Meaning, when the batter is halfway to the first base, his distance from the second base is decreasing at the rate of 11.18 m/sec. --------answer.

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(1.b) At what rate is his distance from third base increasing at the same moment? m/s

Use same figure for the part (1.a) above.

Let S = distance of batter from 3rd base.

S^2 = 90^2 +x^2

2S dS/dt = 2x dx/dt

S dS/dt = x dx/dt

When x = 90/2 = 45m,

S = sqrt[90^2 +45^2] = 100.623 m again

(100.623)dS/dt = 45(25)

dS/dt = 45(25)/100.623 = 11.18 m/sec

Therefore, the distance of the batter from the 3rd base is increasing at the rate of 11.18 m/sec when he is at halfway from the homeplate to the 1st base. --------answer.

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Draw the figure.

It is a right triangle with these:

>>>vertical leg = 1 meter

>>>horizontal leg = x meters .....the distance of the boat's bow from the dock.

>>>hypotenuse = y meters .....the length of the rope from the pulley to the boat's bow.

Given: dy/dt = 1 m/sec

Find: dx/dt when x = 6m.

y^2 = 1^2 +x^2

2y dy/dt = 2x dx/dt

y dy/dt = x dx/dt

y = sqrt[1^2 +x^2] = sqrt[1^2 +6^2] = sqrt(37) m

So,

sqrt(37)*(1) = 6 dx/dt

dx/dt = sqrt(37) /6 = 1.0138 m/sec

Therefore, at the instant the boat's bow is 6 meters from the dock, the boat is approaching the dock at the rate of 1.0138 m/sec. ------answer.