related rates problems!?

Question

1. The surface area of a sphere is decreasing at the constant rate of 3π cm²/sec. At what rate is the volume of the sphere decreasing at the instant its radius is 2 cm?


2. At noon, a ship sails due north from a point P at 8 knots/hr. Another ship, sailing at 12 knots/hr, leaves the same point 1 hour later on a course 60º east of north. How fast is the distance between the ships increasing at 2:00 PM? at 5 PM? Hint: Use law of cosines : a² = b² + c² - 2bc cos A

3. A fly is crawling from left to right along the top of the curve y= 7 - x². A spider waits at the point (4,0). Find the distance between the two insects when they first see each other.

Thanks

Answer 1

1. A = 4πr^2
V = 4/3πr^3
Now dV/dA = dV/dr / dA/dr = 4πr^2 / 8πr = r/2
So dV/dt = dV/dA . dA/dt
= (r/2) (-3π)
= -3π cm^3/sec.

2. Note that knots/hr is not a measure of velocity. Knots are a measure of velocity; 1 knot = 1 nm (nautical mile) / hour. I will assume the given speeds are intended to be in knots.

Let t = time in hours after noon. For t ≥ 1, we have
first ship at 8t nm north of P and the second ship at 12(t-1) nm from P in a direction N60°E. Law of cosines says the distance x between them at time t is given by
x^2 = (8t)^2 + (12(t-1))^2 - 2(8t)(12(t-1)) cos 60°
= 64t^2 + 144t^2 - 288t + 144 - (192t^2 - 192) (1/2)
= 112t^2 - 288t + 240
= 16(7t^2 - 24t + 240)
So x = 4√(7t^2 - 24t + 240)
and dx/dt = 4 (1/2) (7t^2 - 24t + 240)^(-1/2) (14t - 24)
= 4 (7t - 12) / √(7t^2 - 24t + 240)

So, at 2pm the distance is increasing at
4 (14-12) / √(28-48+240) = 8/√220 = 4/√55 ≈ 0.539 knots

and at 5pm the distance is increasing at
4 (35-12) / √(175-120+240) = 92/√295 ≈ 5.36 knots.

Note that you can see from the derivative that the distance reaches a minimum at t = 12/7, i.e. approx. 1:43 pm. So it's not surprising thst the distance is only increasing slowly at 2 pm.

3. I like this one; it's cute. ;-)
If you draw a quick diagram you'll see that what we need is for the tangent line from the curve y = 7-x^2 to pass through (4, 0), which based on my very rough sketch looks to be for x somewhere between 1 and 2. A neater sketch would give a more accurate estimate. ;-)

So, the tangent at the point (x, 7-x^2) is obviously given by dy/dx = -2x. Let's change variables so we don't confuse ourselves when we're trying to get the equation of the line. If the fly is at point (r, 7-r^2) the tangent has gradient -2r. So we need to have the line
y - (7-r^2) = -2r (x-r)
pass through (4, 0), i.e.
0 - (7 - r^2) = -2r (4-r)
<=> -7 + r^2 = -8r + 2r^2
<=> r^2 - 8r + 7 = 0
<=> (r - 1) (r - 7) = 0
<=> r = 1 or 7.
Obviously r = 1 is the one we want, so the fly is at (1, 6) and the distance between them is √(3^2 + 6^2) = 3√5 ≈ 6.708.

Answer 2

1. A= 4pi r^2

dA/dr= 8pi.r
=8 pi.2
= 16 pi

and dr/dA = 1/(16pi)

V= (4/3) pi.r^3

dV/dr = 4pi.r^2
= 4pi.(2^2)
= 16pi
dV/dA = dV/dr x dr/dA

= 16pi . 1/(16pi)
=1

dV/dt = dV/dA x dA/dt

= 1 x 3pi

= 3pi cm^3/sec