Hello!
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
line.
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
derivatives:
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:
http://img45.imageshack.us/my.php?image=for44db9un5.jpg
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:
http://img130.imageshack.us/my.php?image=for52ky4yq0.jpg
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
(approximately):
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
http://www.allaboutcircuits.com/vol_5/chpt_6/6.html
Similar Triangles
http://www.gcseguide.co.uk/similar_triangles.htm
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!
elmarto |