How fast is the tip of her shadow moving when she is 35 ft from the base of the pole?

Question

A street light is at the top of a 14 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 35 ft from the base of the pole?

   _________ ft/sec

Answer 1

pole =14 feet high

woman = 6 feet tall

she moves 4 ft/sec away from the pole

light from top of the pole casts a shadow

how fast is the tip of the shadow moving when she's 35 feet away from the pole?

construct a right triangle. height =14, base =35+x-61.25 ft

b=35,

db/dt=b'=4

construct

a similar right triangle with height 6 feet, base = x

x= distance from the woman to tip of the shadow= 26.25

14/(35+x)= 6/x

14x = 6x+210

x = 210/8= 105/4= 26.25

then you have a related rates problem to find dx/dt=x'

14/(b+x)= 6/x

14x = 6b +6x

8x = 6b

4x =3 b

take the derivative with respect to time

x' = (3/4)b' = (3/4)4) = 3 ft/sec

(b+x)/x = 14/6= 7/3

b+x = 7x/3

7x/3 -3x/3 = b

4x/3 = b

x = 3b/4

x'= 3(4)/4 = 3 ft/sec

the speed of the shadow tip defends on the woman's speed and relative height of the pole to her height

but does not depend on her distance from the pole, the 35 feet is irrelevant

Answer 2

To solve this problem, we can use similar triangles and the concept of related rates.

Let's denote the distance from the woman to the base of the pole as x, and the length of her shadow as y. We are given that the woman is walking away from the pole at a rate of 4 ft/sec, so dx/dt = 4 ft/sec.

We can set up a proportion between the similar triangles formed by the woman, her shadow, and the pole:

(14 ft + 6 ft) / x = 14 ft / y

Simplifying this equation, we have:

20 / x = 14 / y

Now, we can differentiate both sides of the equation with respect to time (t):

d(20 / x) / dt = d(14 / y) / dt

Using the quotient rule, we have:

-20(dx/dt) / x^2 = -14(dy/dt) / y^2

Substituting dx/dt = 4 ft/sec and rearranging the equation, we get:

dy/dt = - (14/y^2) * (x^2/20) * dx/dt

Now, we need to find the value of y when x = 35 ft. Plugging this into the equation, we have:

dy/dt = - (14/y^2) * (35^2/20) * 4 ft/sec

To find y, we can use the similar triangles:

y / x = 14 / (14 + 6)

y / 35 = 14 / 20

Simplifying, we find:

y = (14/20) * 35 = 24.5 ft

Now, we can substitute the values into the equation to find dy/dt:

dy/dt = - (14/(24.5)^2) * (35^2/20) * 4 ft/sec

Calculating this expression, we find:

dy/dt ≈ -2.8 ft/sec

Therefore, the tip of her shadow is moving at approximately -2.8 ft/sec.

Please note that the negative sign indicates that the shadow is getting shorter as the woman moves away from the pole.