Part A)
Here's one way you can draw it out
The points are defined as follows:
Point A = base of lamp
Point B = top of lamp
Point C = base of Bob's feet
Point D = top of Bob's head
Point E = edge of Bob's shadow (assuming the only light source is the lamp)
Based on those points and the given lengths, we know that
AB = 6 (lamp's height)
CD = 2 (Bob's height)
which are both in meters
We also have these unknowns
AC = x
CE = y
DE = z
Triangle BAE is similar to triangle DCE. So we can set up the proportion
AB/AE = CD/CE
AB/(AC+CE) = CD/CE
6/(x+y) = 2/y
6y = 2(x+y)
6y = 2x+2y
6y-2y = 2x+2y-2y
4y = 2x
4y/4 = 2x/4
y = x/2
So you have the correct answer for part A. Nice job.
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Part B)
The first given bit of info states that "Bob... walks away from a lamppost...at a constant rate of 1.5 meters per second"
So this means that x is changing at a rate of 1.5 meters per second; therefore dx/dt = 1.5
dx/dt is the instantaneous speed at a given time t. In this case, it's always 1.5 as it is constant. The expression dx/dt is also positive because he is walking away from the lamp so x will be increasing as t increases.
The goal is to find the expression for dy/dt because we want to know how fast the shadow length is increasing at time t. The value y is the length of the shadow (see diagram in part A)
We'll use the result from part A to get us going. Simply differentiate both sides with respect to t
y = x/2
d/dt [ y ] = d/dt [ x/2 ]
dy/dt = (1/2) * d/dt [ x ]
dy/dt = (1/2) * (dx/dt)
dy/dt = (1/2) * (1.5)
dy/dt = 0.75
So the shadow is increasing at a constant rate of 0.75 meters per second
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Part C)
Go back to the drawing in part A. Focus on triangle DCE. We have a right triangle with legs of 2 and y. The hypotenuse is z.
Using the Pythagorean Theorem, we can say,
a^2 + b^2 = c^2
2^2 + y^2 = z^2
4 + y^2 = z^2
Note: we can solve for z to get
However, it is easier to stick with 4 + y^2 = z^2 when it comes to implicit differentiation.
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Part D)
If Bob is 7 meters from the lamp, then x = 7. So y is
y = x/2
y = 7/2
y = 3.5
Then we can figure out z based on y = 3.5
4 + y^2 = z^2
4 + (3.5)^2 = z^2
16.25 = z^2
z^2 = 16.25
sqrt(z^2) = sqrt(16.25)
z = 4.03112887 (this is approximate)
We'll use the result from part C. Now implicitly differentiate both sides with respect to t
4 + y^2 = z^2
d/dt[4 + y^2] = d/dt[z^2]
d/dt[4] + d/dt[y^2] = d/dt[z^2]
0 + 2y*d/dt[y] = 2z*d/dt[z]
0 + 2y*dy/dt = 2z*dz/dt
2y*dy/dt = 2z*dz/dt
Plug in the values found earlier; isolate dz/dt
2y*dy/dt = 2z*dz/dt
2*(3.5)*(0.75) = 2*4.03112887*dz/dt
5.25 = 8.06225774*dz/dt
dz/dt = 5.25/8.06225774
dz/dt = 0.65118236 (this is approximate)
So the distance from Bob's head to the tip of the shadow is increasing at roughly 0.65118236 meters per second at the exact moment when he is 7 meters from the lamp.
Unlike dx/dt = 1.5 and dy/dt = 0.75, the value of dz/dt is not constant. It depends on y (which is ultimately dependent on x).