Physics Calculations

Question

Each side of a square is increasing at a rate of 8 cm/s. At what rate is the area of the square increasing when the area of the square is 25 cm²?

A plane flying horizontally at an altitude of 1 mi and a speed of 490 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 5 m from the dock? (For this one I tried and tried yet I could not the right questioons,,, I am very sorry

Answer 1

Part I

Let A be area and s be the length of each side,

we have ds/dt = 8 cm/s and when A = 25 cm², s = 5 cm.

dA/dt |@25 cm2 = d (s x s)/dt = d(s²)/dt = (ds²/ds)•(ds/dt) = (2s)•(ds/dt) = (2 x 5 cm) (8 cm/s) = 80 cm²/s // answer

Part II

Draw a right triangle representing the movement of the plane as:

adjacent side = 1 mile, not changing, a, so da/dt = 0 mi/h

opposite side = speed of plane, changing at a rate of 490. mi/h, b, so db/dt = 490 mi/h

hypotenuse = distance btw. the plane and radar station, changing at a rate of dc/dt, that we need to find...

Using Pythagorean Theorem, we have:

a² + b² = c²

take derivative against time on both sides of the equal sign, we have:

d(a² + b²)/dt = dc²/dt

da²/dt + db²/dt = dc²/dt

(2a)da/dt + (2b)db/dt = (2c)dc/dt

(Contact me if you need help in the calculus used here!!!)

since da/dt = 0 mi/h, we simplify and rearranging above equation to:

dc/dt = (b/c)db/dt

put the numbers in:

dc/dt = b/c (490 mi/h)

when c = 2 mi, a is always 1 mi, b must be square root of (c² - a²), √4-1 = √3 mi

dc/dt = √3 mi/2 mi (490 mi/h)

= 424 mi/h // answer

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Part III

Do worry, Hamoton L.!

Let draw the diagram of a right triangle:

Adjacent side: 1 m, the height of the pulley relative to the bow of the boat, a.

Opposite side: the rope, it is moving pulling towards the dock, b.

Hypotenuse: c, the distance between the boat and the dock, that is what we need to find out how fast the boat is approaching the dock when the boat is 5 m away.

Similar to Part II

We have: (2a)da/dt + (2b)db/dt = (2c)dc/dt

da/dt = 0 m/s

dc/dt = 1 m/s               (not db/dt, be careful here! This may be the hindrance for your success!)

so simplify and rearrange to get:

db/dt = (c/b)dc/dt

when b = 5 m, a is 1 m always, and c = √(1² + 5²) = √26 m

So we have:

db/dt = √26 m/5m (1 m/s)

= 1.02 m/s if we take 3 sig. fig. However, there is no reason for us to do that, so let's adhere to the sig. fig. rules, taking 1 s.f.

1 m/s // answer.

if the boat is very close to the dock, let say 0.2 m away, let's find out how fast it approaches...

c = √(1² + 0.2²) = 1.0198 m

db/dt = 1.0198 m/0.2m (1 m/s) = 5.099 m/s, round to 1 s.f. => 5 m/s five times of that of the question being asked.

As you can see, Hamoton, the closer to the dock, the boat is approaching faster! So make sure you slow down the rate of your pull when the boat is closer or else...

From the view point of energy, the closer the boat means the longer you have pulled the boat, the more energy you have given to the boat, or more work done to the boat by you, (some of those energy is spent to cancel friction btw. boat and water...) so, towards the end of your pull, you better do some negative work to the boat to take away some of the energy you put in or otherwise the boat will crash into the dock!!!!

Just like our lives on Earth! We do repetitive things every day: Eat and dump, give and take, up and down... Yet, we progress somehow!

HIH!

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Also, it is important to use calculus as a tool to help you understand the physics, not the other way around!!

If you are not familiar with calculus, a good review from any calculus text will be needed. Most of the needed calculus is provided on the data sheet for, for example, AP Physics C as formulated by the College Board!