Shrek, a 5-ft tall ogre, is walking away from Princess Fiona's castle at the rate of 2ft/sec towards a light source on the ground 60 ft from the castle. How fast is the shadow on the wall of the castle changing when he is walfway between the castle and the light?
2007-01-14 23:19:29 · 8 answers · asked by fra 2 in Science & Mathematics ➔ Mathematics
Easier with a picture, but I'll try to describe.
Imagine a right triangle formed by the light on the ground and the shadow. It has legs 60 and y, the height of the shadow.
There is inside it at any time a similar right triangle, formed by the light and Shrek giving the height. In this case, the legs are x and 5, respectively.
Using similar triangles (right triangles sharing a common angle), you know y/60 = 5/x (ratio of legs), or xy = 300.
Take the implicit derivative of this: ydx + xdy = 0. In the case you care about, x = 30, y = 10, and dx = -2 (minus because since he is walking away, x is getting smaller), so
10*(-2) + 30*dy = 0, dy = 2/3 f/sec
(note: actually, the dx and dy above are more accurately dx/dt and dy/dt, since the derivative is taken with respect to time)
2007-01-15 04:49:46 · answer #1 · answered by sofarsogood 5 · 0⤊ 0⤋
I think my brain exploded when I first read this...
OK, think triangle:
From the castle wall to the light is 60 feet - that is one side of the triangle.
From the ground to the top of Shrek's shadow on the castle wall is another side of the triangle: assuming the ground is flat, this is a RIGHT Triangle. The wall is NOT a triangle side, the shadow is a triangle side - got it?
Shrek's shadow is getting LARGER on the castle wall the closer he gets to the light, so that side of the triangle is always changing.
Halfway between the castle and light is 30 feet.
Your basic right triangle equation is: x squared + y squared = z squared right?
x = distance Shrek has walked
y = height of Shrek's shadow
z squared = x squared + y squared
I THINK that at the beginning
x= 60 feet - distance from light to wall
y = 5 feet - Shrek's height since no shadow has been cast, or you assume the shadow is the same as his height
z = grab a calculator, I'm sleepy
That's all I can do at 4:30 AM :)
This is a rate problem, so you probably have an equation in your notes to get the rest of it.
Good luck and please make this the best answer :)
And you have a good teacher for using cartoon characters, much more interesting, corny but interesting!
2007-01-14 23:34:18 · answer #2 · answered by ? 3 · 0⤊ 1⤋
I don't have much time to help you but I can give you this link, which shows an analogous related rates problem:
http://www.karlscalculus.org/pr8_2-6.html
Give that a try; the problem should be similar enough to know what to do.
2007-01-14 23:30:30 · answer #3 · answered by Puggy 7 · 1⤊ 0⤋
Its a calculus problem. Work out what the equations are then differentiate to find the rate.
2007-01-14 23:23:29 · answer #4 · answered by Anonymous · 1⤊ 1⤋
i need to know the exact position of the light source, whether its convergent or divergent...
2007-01-14 23:31:19 · answer #5 · answered by Death Blade 2 · 0⤊ 1⤋
Is this homwwork if so you do get wuierd homework and I was wondering shouldn't it be in metric?
2007-01-14 23:41:36 · answer #6 · answered by Anonymous · 0⤊ 1⤋
You really have too much time on your hands.
2007-01-14 23:24:13 · answer #7 · answered by trysssa999 3 · 0⤊ 2⤋
2ft/sec??
stagnant??
2007-01-14 23:28:46 · answer #8 · answered by gurl 2 · 0⤊ 0⤋