Saturday, April 14, 2012

Calculus Revisited #8: Related Rates

In our 8th in our 21 blog entry series, we will deal with some problems involving related rates.

Simply put, if you have a function f(x), then f'(x) describes the rate of change at point x.

In physics, you can describe the position of an object given any point in time by the equation x(t). The velocity is found by the first derivative x'(t), and the acceleration is found by the second derivative x''(t).

Position -(derivative)-> Velocity -(derivative)-> Acceleration

Problems
1. A ball is dropped from a building four stories high (48 feet). Let the position equation be x(t) = -1/2 * g * t^2 + 48 where g is the gravitational constant at 32.17405 ft/s^2. When will the ball hit the ground? What will be the velocity and acceleration when the ball hits the ground?

When will the ball hit the ground? How much time will it take for the ball to hit the ground? Solve x(t) = 0.

-1/2 * 32.17405 * t^2 + 48 = 0
-16.087025 * t^2 = -48
t^2 ≈ 2.98377
t ≈ 1.72736 (negative time is nonsensical)

It takes approximately 1.72736 seconds for the ball to reach the ground.

What is the velocity and acceleration when the ball hits the ground?

Velocity: x'(t) = -g * t
x'(1.72736) ≈ -32.17405 * 1.72736 ≈ -55.57617 ft/sec (velocity of the ball moving downward)

Acceleration: x''(t) = -g
The acceleration is 32.17405 ft/sec^2.

2. A rectangular tub 6 feet wide by 4 feet deep is filled with water with its height increasing at 6/5 in/sec. How fast is the volume of water increasing? If the tub is 3 feet high, when will the tub be full?

1 foot = 12 inches

The tub is 72 inches wide by 48 inches deep by 36 inches high. The rate of the increase in height in water is 1.2 in/sec.

Volume of the water = width of the tub * depth of the tub * height of the water

Note that width and depth do not change, only the height.

Then:
V_water(t) = 72 * 48 * h(t)
= 3456 * h(t)
Take the derivative with respect to t:
V_water'(t) = 3456 * h'(t)
The water is rising at 1.2 in/sec. Hence h'(t) = 1.2
V_water'(t) = 3456 * 1.2 = 4147.2

The volume of water is increasing at a rate of 4,147.2 in^3/sec, or 2.4 ft^3/sec.

When will the tub be filled?

The tub will be filled when the height of the water reaches 3 feet, or 36 inches.

distance = rate * time
height of tub = h'(t) * time
36 = 1.2 * t
t = 30

The tub will be full in 30 seconds.

3. Let q(t) = 1/3 * t^3 - t^2 + t coulombs of charge flow through a conducting wire with time (t) being measured in seconds. What is the current after 4 seconds? If there is a 25-amp fuse in the wire, how long will it last?

In electronics, the current of the rate of charge. Current is symbolized by i(t).

Then i(t) = q'(t).

So i(t) = d/dt (1/3*t^3 - t^2 + t) = t^2 - 2t + 1

And i(4) = 16 - 8 + 1 = 9

The current is 9 amp (coulombs/sec).

If there is a 25-amp fuse, how long will it last?

The fuse burns out when the current is 25 amps. So we are solving for t.

i(t) = 25
t^2 - 2t + 1 = 25
t^2 - 2t - 24 = 0
( t - 6 ) ( t + 4 ) = 0

This implies that t = 6 (negative time is nonsensical).

The fuse lasts for 6 seconds.

This concludes Part 8 of this 21 part series. Next time, we will solve non-linear equations.

In good calculation, Eddie.

This blog is property of Edward Shore. © 2012

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