Expert: Paul Klarreich - 11/21/2005

Question
1) the minute hand of a certain clock is 4 in. long. starting from the moment when the hand is pointing straight up, how fast is the area of the sector that is swept out by the hand increasing at any instant during the next revolution of the hand?
2) a 13-ft ladder is leaning against a wall. if the top of the ladder slips down the wall at a rate of 2 ft/s, how fast will the foot be moving away from the wall when the top is 5 ft above the ground?

Answer
Hi, Amy, once again.
[SORRY - CLICKED THE WRONG SPOT BEFORE.]

Question:  1) the minute hand of a certain clock is 4 in. long. starting from the moment when the hand is pointing straight up, how fast is the area of the sector that is swept out by the hand increasing at any instant during the next revolution of the hand?

You really need a mathematical handbook.  I will get you started and then you can do the rest.

Variables:
A = sector area.
theta = angle of rotation in RADIANS.

Relation:  Area formula of a sector.  YOU LOOK IT UP.
OK, OK, then.  

The area of a sector is  A = theta r^2/2
(You could have guessed that.  

Rates:
dA/dt = increase of sector area to be found.
dtheta/dt = rate of increase of the angle = pi/6

Why pi/6?  This can be computed by the basic rule that in 12 hours, theta will increase by 2 pi radians, so the rate is 2 pi/12 = pi/6 radians per hour.

OK.  Just differentiate that formula, put in your dtheta/dt = pi/6, and you're on your way.  Note that r = 4 is not a variable.
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2) a 13-ft ladder is leaning against a wall. if the top of the ladder slips down the wall at a rate of 2 ft/s, how fast will the foot be moving away from the wall when the top is 5 ft above the ground?

h = height of ladder
x = distance of foot to wall.

Rates:
dh/dt = rate ladder slips = -2
dx/dt = rate foot moves away, to be found.

Relation:
DRAW THE DIAGRAM, LOOK AT THE FIGURE, DECIDE WHAT THE SITUATION IS.

If you haven't guessed by now, this is a right triangle, and x^2 + h^2 = 13^2