By: Courtney Kent
Carpentry/PDT
A carpenter is cutting down a 200 foot tree
on a construction site; the sun is directly
overhead. At the moment when the tree
makes an angle of 30° with the horizontal, its
shadow is lengthening at the rate of 50
feet/sec. How fast is the angle changing at
that moment?
Step1: Draw a diagram
portraying the problem.
Step 2: Fill in the diagram
with the given information.
The rate at which the
shadow is lengthening is
equal to the derivative of
side A. The tree is a
constant , it remains 200ft.
The tree is equal to side C.
The angle created when
side A and C meet is 30°.
What the problem is trying
to ask for is dθ/dt.
B
30°
50ft/sec
A
cos(θ)=a/c
(-sin θ)(dθ/dt)= (c)(da/dt)- (a)(dc/dt)
(c²)
(-sin θ)(dθ/dt)= (c)(da/dt)
(c²)
(-sin θ)(dθ/dt)= 200(50)
(200²)
(-sin θ)(dθ/dt)= 1
4
(-sin (30))(dθ/dt)= 1
4
(0.988)(dθ/dt)= 1
4
(0.988)(dθ/dt)= (0.25)
(0.988)
(0.988)
(dθ/dt)= 0.253°
sec
Step3:Use the derivative of
cosine to find the missing
information. Chain rule
(first derivative time the derivative of what is on the inside)
is used in this process as
well as the quotient rule
(low dihigh- high dilow/ the square of what is below)
The second part of the
chain rule(A•dc/dt) cancels
because A is a constant
and therefore does not
have a derivative.
Step 4:Solve the equation
for dθ/dt.
At the moment when the tree makes an angle
of 30° with the horizontal, and its shadow is
lengthening at the rate of 50 feet/sec the
angle changing 0.253° per second at that
moment.
A construction worker pulls a five meter
plank up the side of a building under
construction by means of a rope tied to one
end of the plank. Assume the opposite end of
the plank follows a path perpendicular to the
wall of the building and the worker pulls the
rope at a rate of 0.15 meters per second.
How fast is the end of the plank sliding along
the ground when it is 2.5 meters from the
wall of the building?
Step1: Draw a diagram
portraying the problem.
Step 2: Fill in the diagram with
the given information. The
distance from the building, 2.5
meters, is side A. The rate at
which the rope is being pulled
is 0.15 meters per second.
This is the derivative of side B
The plank is equal to side C.
The plank is a constant , it
remains 5 meters. What the
problem is trying to ask for is
the derivative of side A.
0.15m/sec
B
2.5 meters
A
(a)²+ (b)²= (c)²
(2.5)²+ (b)²= (5)²
6.25 + (b)²= 25
-6.25
-6.25
(sqrt)(b)²= (sqrt of 18.75)
b = 4.330127
(a)²+ (b)²= (c)²
(2a)(da/dt)+(2b)(db/dt)=(2c)(dc/dt)
(2(2.5)(da/dt)+(2(4.330127)(0.15)=(2(5)(0)
5(da/dt) + 1.3 = 0
- 1.3
- 1.3
5(da/dt)= -1.3
5
5
(da/dt)= -0.26 m/sec
Step3:Use the
Pythagorean Theorem to find
side B.
Step4:Use the derivative of
the Pythagorean Theorem to
find the missing information.
Chain rule
(first derivative time the derivative of what is on the inside)
is used in this process. The C
part of the
Pythagorean Theorem when
solving for a derivative
cancels out. C is a constant
and therefore does not have a
derivative.
Step 5:Solve the equation for
dA/dt.
The end of the plank sliding along the ground
at a decreasing rate of 0.26 meters per
second when it is 2.5 meters from the wall of
the building.