SOLUTION: Any help is much Appreciated!!!! Although this is an albera site, i have calculus problems that involve alot of algrebra as well. I'm lost & have alot of homework due on monday mo

Question 188620: Any help is much Appreciated!!!! Although this is an albera site, i have calculus problems that involve alot of algrebra as well. I'm lost & have alot of homework due on monday morning. I started drawing pictures to help me, but then got confused about if I drew the picture right and don't know what to do after that.
Story Problems:
1.) A shallow barge is being drawn towards a dock by a taut (not swaggering) cable attached to the top of a dock 6ft above the deck of the barge. When the barge is 24ft away from the bottom(water level) of the dock, it is moving in the water at a speed of 3ft per second towards the dock. At this time, how fast is the cable being pulled in?

2.) Two street lights of hight L are a distance D apart. The light at the top of one is functioning but the other is being worked on by a repair person. If the repair person drops a wrench from rest at the top of the lamp pole, how fast is the wrench's shadow(made in the light of the other street lanmp) moving horizontally along the ground when it is a height, h, above the ground?
[HINT: in the absence of air resistance, conservation of energy requires that (1/2)v^2 + gh = gL where v is the downward velocity of the wrench, and g is a constant, known as the acceleration of gravity. Answer should be a function of g, L, D, and h .]
3.) Two points are marked on the surface of a spherical balloon. One point is marked at the 'North Pole', while the other is on the 'Equator'.
a.) In terms of the sphere's radius, r, what is the distance, s, between the two points as measured on the surface of the sphere?
(i.e. What is the shortest length of strong on can lay on the sphere between the two points?)
b.) In terms of the sphere's radius, r, what is the distance, D, between the two points as measured in three-dimensional space?
(i.e. What is the shortest length one can Position within the sphere between the two points?)

c.) Air is pumped into the ballon in such a way that the ballon's volume is increasing at a constant rate of 3000(cm^3/min)
When r= 10.0 cm, how fast is, s, increasing?
d.) When r= 10.0 cm, how fast is D increasing?

THANK YOU TONS FOR ANY HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
1.) A shallow barge is being drawn towards a dock by a taut (not swaggering) cable attached to the top of a dock 6ft above the deck of the barge. When the barge is 24ft away from the bottom(water level) of the dock, it is moving in the water at a speed of 3ft per second towards the dock. At this time, how fast is the cable being pulled in?
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This makes a right triangle with sides A = 6 and B = 24 and hypotenuse C.
Side A = (C^2 - A^2)^(1/2)
Differentiate wrt to time
dA/dt = (1/2)*((C^2 - A^2)^(-1/2)*2C dC/dt = 3 ft/sec
Solve for dC/dt = 12/sqrt(17) = 2.91...ft/sec
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c.) Air is pumped into the ballon in such a way that the ballon's volume is increasing at a constant rate of 3000(cm^3/min)
When r= 10.0 cm, how fast is, s, increasing?
d.) When r= 10.0 cm, how fast is D increasing?

Differentiate wrt time

dV/dt = 3000 cc/min

dr/dt = 7.5/pi change in radius
D is the hypotenuse where r is the 2 legs, or r*sqrt(2).
dD/dt = 7.5*sqrt(2)/pi (it increases linearly with r)
dD/dt = ~3.3762 cm/min
--------------------
When r= 10.0 cm, how fast is, s, increasing?
s = 1/4 of the circumference = pi*r/2 = 5pi
ds/dt = (pi/2)*dr/dt
ds/dt = 3.75 cm/min
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I'll do the rest this weekend. email me via the thank you note if you have questions.
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2.) Two street lights of hight L are a distance D apart. The light at the top of one is functioning but the other is being worked on by a repair person. If the repair person drops a wrench from rest at the top of the lamp pole, how fast is the wrench's shadow(made in the light of the other street lanmp) moving horizontally along the ground when it is a height, h, above the ground?
[HINT: in the absence of air resistance, conservation of energy requires that (1/2)v^2 + gh = gL where v is the downward velocity of the wrench, and g is a constant, known as the acceleration of gravity. Answer should be a function of g, L, D, and h .]
I don't know about that conservation of energy. The speed of a falling object is v = gt, and the distance it falls is (gt^2)/2. The height of the wrench at t seconds is L - (gt^2)/2, and it's falling at gt (usually in feet per second).
Assume both the light and wrench are points.
To find the distance s from the light to the shadow, use the 2 similar triangles, the larger one from the light, the ground and the shadow, and
the smaller one: The light, D, and the distance the wrench has fallen.
L/s = (gt^2/2)/D
s = LD/(gt^2/2) = 2LD(t^-2)/g
ds/dt = (2LD/g)*(-2t^-3)
ds/dt = -4LD/(gt^3) (negative because the distance is decreasing)
------------------
This is the speed of the shadow as a function of time, no h.
h is also a function of time, it's L - (gt^2)/2
h can be subbed into the eqn, but there's no advantage to that.
ds/dt = -4LD/(L-h)t
Doing any calculations with this one is more tedious, and the relation between the speed of the shadow and time is less clear.
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