There's these two word problems I don't understand.

1. A man 6 ft tall walks at a rate of 5 ft per second toward a streetlight that is 30 feet high. The man's 3-foot-tall child follows at the same speed, but 10 feet behind the man.
a) Suppose the man is 90 feet from the streetlight. Show that the man's shadow extends beyond the child's shadow.
b) Suppose the man is 60 feet from the streetlight. Show that the man's shadow extends beyond the child's shadow.
c) Determine the distance d from the man to the streetlight at which tips of the two shadows are exactly the same distance from the streetlight.
d) Determine how fast the tip of the man's shadow is moving as a function of x, the distance between the man and the streetlight. Discuss the continuity of this shadow speed function.

2. A particle is moving along the graph of y= the third cube root of x. When x=8, the y-component of the position of the particle is increasing at the rate of 1 centimeter per second.
a) How fast is the x-component changing at this moment?
b)How fast is the distance from the origin changing at this moment?
c) How fast is the angle of inclination theta changing at this moment?

Rats. I made a typo near the top of #2.

dy/dt = 1/(3∛x^2) dx/dt
(a) at x=8,
1 = 1/(3*4) dx/dt
so, dx/dt = 12 at x=8, not 6.

You will have to use that in the other parts.

Also, in part (c) I made a typo. It should be

sec^2θ dθ/dt = (-2/3)x^(-5/3) dx/dt, so
dθ/dt = -36/160

(a)

If the man is x feet from the light pole, and his shadow has length m, then using similar triangles.

(x+m)/30 = m/6

In like wise, if the child's shadow has length c, then

(x+10+c)/30 = c/3

So, we need to show that x+m > x+10+c when x=90

90+m=5m
90+10+c = 10c

m=90/4 = 22.5
c=100/9 = 11.1

90+22.5 = 111.5
100+11.1 = 111.1 < 111.5
So the man's shadow extends beyond the child's.

(b) repeat with x=60
I suspect a typo.

(c) should not be too hard

(d) Since (x+m)/30 = m/6
x+m = 5m
x = 4m
dx/dt = 4 dm/dt
So the shadow is shrinking at 1/4 the man's walking speed.

#2
y = ∛x
dy/dt = 1/(3∛x^2) dx/dt
(a) at x=8,
1 = 1/(3*2) dx/dt

(b) the distance z is
z^2 = x^2+y^2 = x^2 + ∛x^2
at x=8, the point (8,2) is √68 from the origin
z dz/dt = x dx/dt + y dy/dt
√68 dz/dt = 8*6 + 2*1
dz/dt = 50/√68 = 25/√17

(c) tanθ = y/x = ∛x/x = x^(-2/3)
sec^2θ dθ/dt = (-2/3)x^(-5/3)
at x=8, tanθ = 2/8 = 1/3, so sec^2θ = 1+tan^2θ = 10/9
10/9 dθ/dt = (-2/3)(1/32)
dθ/dt = -3/160

Sanity check: The curve's slope is less than 1, so as x increases, the angle is decreasing.