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Q: 3 calculus problems ( Answered 5 out of 5 stars,   4 Comments )
Subject: 3 calculus problems
Category: Science > Math
Asked by: jeremit-ga
List Price: $100.00
Posted: 25 Jul 2006 11:14 PDT
Expires: 24 Aug 2006 11:14 PDT
Question ID: 749386
**I need these answered before Wednesday July 26 3:00pm**

1. A man 6 feet tall walks at a rate of 5 feet per second away from a light that 
   is 15 feet above the ground. When he 10 feet from the base of the light,

   a) At what rate is the tip of his shadow moving?
   b) At what rae is the length of his shadow changing?


2. A fish is reeled in at a rate of 1 foot per second from a point 10 feet above 
   the water . At what rate is the angle between the line and the water changing 
   when there is a total of 25 feet of line out?


3. A ball is dropped from a height of 20 meters, 12 meters away from
the top of a 20 meter lamppost. The ball's shadow, caused by the light
 at the top of the lamppost, is moving along the level ground. How
fast is the shadow moving 1 second after the ball is released?

Subject: Re: 3 calculus problems
Answered By: elmarto-ga on 26 Jul 2006 09:06 PDT
Rated:5 out of 5 stars
Here are your answers.

QUestion 1
Let's call x to the distance (across the ground) from the lamppost to
the man and s to the length of the shadow. As you can see in your
image, there are are two triangles here:

1) The big triangle whose sides are the lamppost, x + s (the distance
from the lamppost to  the tip of the shadow) and the whole dashed

2) The small triangle, contained in the big one, whose sides are the
man, his shadow, and the the part of the dashed line that starts at
his head and ends in the ground.

Clearly, these are similar triangles. As such, the ratio between their
sides should be equal. Therefore, we get the following equation:

6/15 = s/(x+s)

That is, the ratio between the "vertical" sides of the triangles (6
for the small one and 15 for the big one) must be equal to the ratio
between the "horizontal" sides (s for the small one and (x+s) for the
big one). Using this equation, we can get the length of the shadow as
a function of the distance of the man from the lamppost:

6(x+s) = 15s
6x = 9s
s = (2/3)x

We want to find ds/dt (the rate of change of the length of the
shadow). Using the above equation:

ds/dt = (2/3)*(dx/dt)

Finally, we know that dx/dt = 5 (the distance between the man and the
lamppost is increasing at 5 ft per second). Therefore,

ds/dt = (2/3)*5 = 10/3

The length of the shadow is thus growing at 10/3 ft per second. Notice
that this rate of change is independent of the current distance from
the man to the post (10 ft)

The rate of change of the tip of the shadow is easy now. The distance
between the tip and the lamppost is (x + s). Using a basic property of

d(x + s) / dt = dx/dt + ds/dt = 5 + 10/3 = 25/3

The tip of the shadow is thus moving away from the lamppost at 25/3 ft per second.

Question 2
In order to solve this, we follow the same strategy as before: we want
to write the variable for which we want to find the rate of change as
a function of other known variables.

In this case, notice from your graph that there is a clear right
triangle. I've marked it in your graph:

Let's call A to the angle. Since the opposite side of this angle is 10
ft, and the hypotenuse of the triangle is x (the length of the fishing
line), we get that:

sin(A) = 10/x

We take derivative with respect to t of both sides of this equation:

d[sin(A)]/dt = d(10/x)/dt

Using the chain rule for derivatives, we get:

cos(A)*(dA/dt) = -(10/x^2)*(dx/dt)
[the ^ symbol means "to the power of", so we have an x squared in the
denominator on the right side]

Isolating dA/dt, which is what we are interested in:

dA/dt = -(10/x^2)*(dx/dt)/cos(A)

Now, recall that we're told that the line is being reeled at 1 ft per
second. Therefore, x is getting smaller at a rate of 1 ft/second. We
thus have that dx/dt = -1. So the equation becomes:

dA/dt = 10/[x^2 * cos(A)]

Now, the length of the line (x) is currently 25. So now we just need
to find A when x is 25. We do this using basic trigonometry. Since
sin(A) = 10/25, then A = arcsin(10/25). So we get the final equation:

dA/dt = 10/[x^2 * cos(A)]
dA/dt = 10/[25^2 * cos(arcsin(10/25))]
dA/dt = 0.01745 rad

We've thus found that, when there is a total of 25 feet of line out
and its being reeled at 1 ft per second, the angle is growing at
0.01745 radians per second.

Question 3
This problem can be solved in a very similar fashion to the first one.
Once again, we'll use like triangles. Check the graph below:

The blue and red triangles are clearly similar triangles. Let's call x
to the height of the ball (which would be the height of the blue
triangle) and s to the base of the blue triangle (so that 12+s is the
distance between the lamppost and the shadow of the ball). Since these
are similar triangles, we get that:

x/20 = s/(12+s)

Isolating s:

(12+s)x = 20s
12x + sx = 20s
12x = s(20-x)
s = 12x/(20-x)

So now take the derivative with respect to t:

ds/dt = d[12x/(20-x)]

I assume now that you are familiar with the rules for finding the
derivative of a quotient. If you're not, please request clarification
and I'll gladly explain. We get that:

ds/dt = [12(dx/dt)(20-x) + 12x(dx/dt)]/(20-x)^2

So we need to know x and dx/dt. The formula for the height (in meters)
of an object dropped from height h0 after t seconds is known to be

h0 - 4.9*t^2

So, we know that the initial height of the ball (h0) was 20 meters,
and that time elapsed since the ball was dropped (t) was 1 second.
Therefore, the height of the ball after 1 second would be:

20 - 4.9*(1^2) = 20 - 4.9 = 15.1 meters

We've thus found that x = 15.1. Now we need to find dx/dt. Clearly,
this is the speed at which the ball drops. The acceleration of gravity
is 9.8 m/s^2, meaning that, when you drop an object, its speed
increases by 9.8 m/s each second. Since the speed of the ball is
initially zero, and one second has passed, then the speed of the ball
after that one second must be 9.8 m/s. Since the height of the ball is
getting smaller (because it's dropping), we conclude that

dx/dt = -9.8.

Now we know that x = 15.1 and dx/dt = -9.8. We plug these values into
our derivative:

ds/dt = [12*(-9.8)*4.9 + 12*15.1*(-9.8)]/(4.9^2)
ds/dt = -97.96

We've found that, after 1 second, the shadow moves towards the ball at
a speed of 97.96 meters per second.

Additional Sources
Rules for derivatives

Similar Triangles

Google search terms
"similar triangles"
derivatives rules

I hope this helps! If you have any doubt regarding my answer, please
don't hesitate to request clarification before rating it. Otherwise, I
await your rating and final comments.

Best wishes!
jeremit-ga rated this answer:5 out of 5 stars and gave an additional tip of: $3.00
Thank you for answering this!!!

Subject: Re: 3 calculus problems
From: cynthia-ga on 25 Jul 2006 13:38 PDT
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Subject: Re: 3 calculus problems
From: jeremit-ga on 26 Jul 2006 05:48 PDT
To the poster above me, im actually just a calc 1 student and
optimization problems of this complexity for the most part are beyond
the scope of my ability as well as many others in my current 8-week
class. Nothing like this would be on an exam or done for a grade in
general, but my instructor throws problems at us from time to time to
see if we can work together in a group setting to solve them.....but
they are not graded. We haven't even discussed integrals as of yet,
but will very soon. He might go over these during the next class
session (hence the time limit), but knowing him there definitely is no
guarantee. Your response pretty much guarantees that they will not get
answered so i'll just cancel them.
Subject: Re: 3 calculus problems
From: cynthia-ga on 27 Jul 2006 17:53 PDT
Got it --in the future it would help you to get an answer if you
explained that in the question (or a clarification) so the researchers
don't jump to conclusions!  As it stood, it sure *looked* like
Subject: Re: 3 calculus problems
From: elmarto-ga on 01 Aug 2006 15:47 PDT
jeremit: I'm glad you liked my answer. Thank you very much for your
comments and tip!

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