Calculus related rates HELP!?

Question

A street light is at the top of a 13 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 30 ft from the base of the pole?

can anyone help me work it out step by steps? ty very much

Answer 1

Step 1 - Draw a picture!!!

Yahoo doesn't allow me to draw very good pictures, but here's the gist...

|
|
|----------I
|----------I--------
~~y~~~
~~~~~x~~~~~

The vertical leg is the street light (13 ft) and the horizontal leg is where the tip of the woman's shadow is (x ft). The I is where the woman is standing at any given instant (y ft).

We can draw a smaller right triangle... the top of the woman's head to the tip of her shadow.

Recap:

Bigger Triangle
Vertical Leg = 13 ft
Horizontal leg = x ft

Smaller Triangle
Vertical Leg = 6 ft
Horizontal Leg = (x - y) ft

Notice that the two triangles are similar...

We can set up a ratio...

13 / 6 = x / (x - y)

Cross multiply

13x - 13y = 6x

Differentiate

13 dx / dt - 13 dy/dt = 6 dx/dt

We know that dy/dt represents the woman's speed (5 ft/s) and dx/dt is what we're looking for (the shadow's speed).

13 dx/dt -13 dy/dt = 6 dx/dt
13 dx/dt - 13 *5 = 6 dx/dt
-65 = 6 dx/dt - 13dx/dt
-65 = -7 dx/dt
65/7 = dx/dt


Notice that the fact that she is 30 ft from the base doesn't matter... Her shadow is growing at a constant rate as she walks at a constant rate.

Answer 2

Always draw a picture. Put the street light pole at one end of the picture, and let A be the point at which the pole touches the ground and B, where the bulb is. At 30 feet away, let the woman be represented by C-D, where D is where she touches the ground and C is the top of her head.
Draw a line B-C-E, where E is at the ground. Then D-E is the shadow when the women is at 30 feet. She is moving at 5 ft/sec from D to E. The question is how far will E be moving?

EAB is a right triangle, which is similar to EDC. By the properties of similarity,
...... DE/DC = AE/AB
Now AE = AD (the distance to the women ) .+
DE)the length of the shadow.
Breaking up AE, DE/DC=AD/AB + DE/AB.
putting in some numbers, DE/6 = AD/13+ DE/13
Sorting out variables, DE/6-DE/13= 7DE/78= AD/13 and 91/78 DE or 7/6 DE=AD
We know d(AD)/dt =5, so we should be able to find d(DE)dt = 30/7 or 4.29 ft/sec.

Answer 3

First you need an equation to relate the distance from the pole to the length of her shadow. Call d the distance between the woman and the pole and call s the length of the shadow. With similar triangles (assuming you have a correct picture) you will find that,

6/13=s/(s+d)
6d=7s

now we differentiate

6d'=7s'

Since we know that d' is 5,

6*5=7s'
s'=30/7

I hope this helps.

Answer 4

Draw a diagram with the pole on one side and the girl in the middle. Draw a line for the tip of the pole over the tip of the girl's head, to the ground. You will have a triangle. Let the pole be at zero, the girl at x, and the intersection on the ground at y. You want to calculate the speed of y, that is dy/dt. If the height of the pole is h and the height of the girl is g, it is easy to see that h/y=g/(y-x), by similar triangles. Solving for y, you get y = h/(h-g) x. Taking d/dt of both sides, dy/dt = h/(h-g) dx/dt. Putting in numbers, dy/dt = 13/(13-6) * 5 =65/7 fps.

Answer 5

[a] floor section = circumference * height = 2? r * h height = 10cm + 0.1cm * t h = 10 + 0.1t A = 10? * (10 + 0.1t) = 100? + ? t dA/dt = value of exchange of section with relation to time dA/dt = (100? + ? t)' = ? cm^2/sec [b] volume = component to base * height = ? r^2 * (10 + 0.1t) V = 25? (10 + 0.1t) V = 250? + 2.5? t dV/dt = 2.5? cm^3/sec desire this facilitates!

Answer 6

I think this link may be very helpful:

http://en.wikipedia.org/wiki/Related_rates

I did this in the beginning of my Calculus class this year but since no related rates exist on the BC exam I didn't even bother to study it.

Answer 7

LOL. I'm stuck on the exact same problem. It's quesiton #9.