Differentiation word problem -1 and 2?

Question

1) Oil spilled from a ruptured tanker spreads in a circle at a constant rate of 6 miles^2/hr. How fast is the radius of the spill increasing when the area is 9mi^2?

2) a 17 ft ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of 5 ft/sec., how fast will the top of the ladder be moving down the wall when it is 8 ft above the ground?

If anybody can answer these questions and explain step by step in detail, you are a genius!!

Answer 1

1) A = pi x r^2
differentiating both sides, we have
dA/dt = pi x 2r dr/dt

from the first equation, when A = 9mi^2, r = 3/sqrt(pi)
dA/dt is constantly 6mi^2/hr, so the second equation will become

6 = pi x 2(3/sqrt(pi)) dr/dt = 6sqrt(pi) dr/dt

dr/dt = 1/sqrt(pi) or sqrt(pi)/pi mi/hr. :)

2) pythagorean's relation tells us that
x^2 + y^2 = 17^2
differentiating, we have
2x dx/dt + 2y dy/dt = 0

from the first equation, when y = 8ft, x = 15 ft.
(8^2 + 15^2 = 17^2)

dx/dt is constantly 5ft/sec, so plugging in the values in the second equation, we have
2(15)(5) + 2(8)dy/dt = 0
dy/dt = -150/16 = -75/8 ft/sec

the value is negative because it's sliding down the wall :)

Answer 2

1. The area is pi r^2. If the rate of area spread is 6 square miles per hour, then dA/dt = 6 = 2r dr/dt, and if the area is 9, the radius is 3/sqrt(pi). Putting this together, we have 3/sqrt(pi) dr/dt = 3, or dr/dt = 1/sqrt(pi).

2. We have x^2 + y^2 = 289. Differentiating with respect to time, we have 2x dx/dt + 2y dy/dt = 0. Plugging in, 2x (5) + 2 (8) dy/dt = 0, and x = sqrt(289-64) = 15. Then dy/dt = -30 (5) /16, or about -9 feet per second.

Answer 3

1) Area of spill = A = pi * r^2
When area is 9 sq. miles, we have 9 = pi * r^2
or r^2 = 9/pi
or r = sqrt(9/pi) = 1.6925 miles

dA/dt is a constant 6 sq.miles/hour
But dA/dt = d(pi * r^2)/dt
= pi * (2r) ( dr/dt) = 6
So dr/dt = 6 / (2 * pi * r)
= 3/ (pi * r)
When r = 1.6925 miles, we get
dr/dt = 3/( 3.1418 * 1.6925)
= 0.564 miles/hour

Q2. If y is the height, and x is the distance of the foot of the ladder from the wall, you need to express dy/dt in terms of dx/dt, which is a constant 5 ft/sec.

Use the fact that x^2 + y^2 = 17^2 to solve for x when y=8. Also differentiate both sides of this equation wrt t , and simplify it to get dy/dt as a function of x, y and dx/dt; you know the values for these when y=8.

Answer 4

Related rates!

I'll do problem 1, you do problem 2. The concept is basically the same, except you're dealing with a triangle in stead of a circle (thing pythagorean theorem, etc).

So we have a circle.

We are told that the oil spreads at a certain rate (6 mil^2/hr).

What this means is that the circles area is increasing
at a rate of 6 miles ^2/hour.

So we have the formula for area:

A=pi(r^2)

and are told that
dA/dt=6 (change in area with respect to time is 6)

Now we differentiate.

dA/dt=2pi(r)dr/dt

Now, when the area is 9, the radius is

9=pi(r^2)
r=3/sqrt(pi)

So now again

dA/dt=2pi(r)dr/dt=2pi(3/sqrt(pi))dr/dt=6

dr/dt=1/sqrt(pi) miles/hour

Answer 5

you do not decide for calculus to unravel this. a million. the quantity is 15000 gals = 1705 ft^3 quantity of a sphere = 4/3 * pi * R^3 resolve that to get R = 7.40-one ft whilst that's a million/2 finished, the section is pi * R^ 2 = 172.6 ft^2 The flowrate is a hundred gal/min = 11.36 ft^3/min Dividing those, the exchange of intensity each and every minute is 0.0658 ft. 2. whilst that's a million/2 finished, it contains 7500 gallons of water and there are 7500 gallons to circulate. So at a hundred gallons in step with minute, you have have been given 75 minutes to circulate.

Answer 6

1) A = πr^2
r = √(A/π)
dA/dt = 2πrdr/dt
dr/dt = (dA/dt)/(2π√(A/π))
dr/dt = (6)/(2π√(9/π))
dr/dt = 0.56419 mi/hr

2)y^2 = L^2 - x^2
x = √(L^2 - y^2)
2ydy/dt = - 2xdx/dt
dy/dt = - (x/y)dx/dt
dy/dt = - (√(L^2 - y^2)/y)dx/dt
dy/dt = - (√(17^2 - 8^2)/8)5
dy/dt = - 9.375 ft/sec