Calculus AB: Applications of the Derivative

Problem : David and Angela start at the same point. At time t = 0 , Angela starts running 30ft/sec north, while David starts running 40ft/sec east. At what rate is the distance between them increasing when they are 100 feet apart?

Solution for Problem 1 >>

1) The particular information here is that they are 100 feet apart. This information should be set aside until the last step.
2) In the diagram, let x(t) =distance traveled by David east let y(t) =distance traveled by Angela north, and the let z(t) be the distance between them

3) Here we have x'(t) and y'(t) , and we want to find z'(t)
4) x(t) + y(t) = z(t)
5) 2x(t) +2y(t) = 2z(t)
6) When z(t) = 100 , x(t) = 80 and y(t) = 60

2(80)(40) + 2(60)(30)	=	2(100)
	=	50 feet per second

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Problem : Sophia is sitting on the ground 10 feet from the spot where a hot air balloon is about to land. She is watching the balloon as it travels at a steady rate of 20 feet per second towards the ground. If θ is the angle between the ground and her line of sight to the balloon, at what rate is this angle changing at the instant the balloon hits the ground?

Solution for Problem 2 >>

The particular information is that the balloon has hit the ground, i.e. θ(t) = 0 . This should be set aside for now. The graph is as follows, with h(t) representing the height of the balloon from the ground.

We have . We want to find .

		tan θ(t) =
		sec2 θ(t) =
		Rewrite sec as the reciprocal of the cos:
		=
		Now plug in the particular values:
		= (- 20)
		= - 2 radians per second

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Problem : Indy is 6 feet tall and is walking at a rate of 3 feet per second towards a lamppost that is 18 feet tall. At what rate is his shadow due to the lamppost shortening when he is 6 feet from the base of the lamppost?

Solution for Problem 3 >>

The particular information is that he is 6 feet from the lamppost. Let x be the length of the shadow, and let y be the distance between Indy and the lamppost.

We have , and we wish to find .

		By similar triangles,
		=
		+ =
		= - 3, since this quantity is decreasing.
		-3 =
		= - feet per second

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Problem : Water is being poured into an inverted cone (has the point at the bottom) at the rate of 4 cubic centimeters per second. The cone has a maximum radius of 6cm and a height of 30 cm. At what rate is the height increasing when the height is 3cm?

Solution for Problem 4 >>

The particular information is that height=5cm.

We have ; we want .

V(t) = Π r(t) h(t)

Taking the derivative now would generate the term which we do not have any information about. To avoid this problem, we eliminate r(t) from the equation by using similar triangles to generate the following relation:

=

so,

r = h

substituting this into the equation yields

V(t)		= Π h(t)3
		= h(t)2

Using h = 5 ,

4		= (5)2
		= cm per second

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