b*****n
发帖数: 143 | 1 看起来好象很简单：
Suppose there is an infinite straight beach and there is a lighthouse 1 mile
offshore. The light rotates at 1 revolution per minute. How fast is the
image of the beam on the beach, i.e. the "white dot", moving along the beach
when that white dot is exactly 3 miles from the lighthouse?
我觉得答案是２＊Pi miles/minute，不对，标准答案不知道。
先谢谢诸位。 |
n*******e
发帖数: 55 | 2 |\
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L
L=tan(A)->dL/dt=(1+tan^2(A))dA/dt=(1+9)*2*pi per minute=20*pi per minute
No offense but as you mentioned, this is a easy one. Anyone with a little
bit knowledge of trig derivative should be able solve it.
mile
beach
【在 b*****n 的大作中提到】
: 看起来好象很简单：
: Suppose there is an infinite straight beach and there is a lighthouse 1 mile
: offshore. The light rotates at 1 revolution per minute. How fast is the
: image of the beam on the beach, i.e. the "white dot", moving along the beach
: when that white dot is exactly 3 miles from the lighthouse?
: 我觉得答案是２＊Pi miles/minute，不对，标准答案不知道。
: 先谢谢诸位。
|
b*****n
发帖数: 143 | 3 Thanks for your help. But according to your formula, shouldn't tan^2(A) be 1
/8 instead of 9? Maybe I missed something?
This is my solution: when the beam image is 3 miles from the light house,
its velocity at that point is:
V = 3*2*Pi miles/min
Because this vector is not along the beach, the component that is parallel
to the beach should be:
V(b) = V * 1/3
where 1/3 is cosine of the angle between the beach and the velocity.
Thanks. |
n****y
发帖数: 28 | 4 You can find a lot of similar problems in some calculus book. |
b*****n
发帖数: 143 | 5 Why it is 6*pi divided by cos(A)? I keep thinking that it should be multiply
. Thanks.
So |
n****y
发帖数: 28 | 6 lol, I didn't notice that.
is
So |
