Related rates help needed

[sigma]

I need help with the following related rate questions. If somebody would be as so kind to provide a solution as I have been trying to get help with them for the last week with no response from who I seek help from usually. Please help! I know the answers but need a solution so here they are. If the diagrams could be explained, that would be greatly appreciated.

Question 1. Jim, who is 180 cm tall, is walking towards a lamp-post which is 3 meters high. The lamp casts a shadow behind him. He notices that his shadow gets shorter as he moves closer to the lamp. He is walking at 2.4 meters per second.

a)When he is 2 meters from the lamp-post, how fast is the length of his shadow decreasing?

b)How fast is the tip of his shadow moving?

The answer to a) is 3.6 m/s and the answer to b) is 6 m/s but solution needed.


Question 2. A person 2 m tall walks towards a lamp-post on level ground at a rate of 0.5 m/sec. The lamp on the post is 5 m high. How fast is the length of the person's shadow decreasing when the person is 3 m from the post?

The answer to this one is -1/3 m/s.

All the help is greatly appreciated! I have a midterm exam next Friday and need these mastered by then.

 

[soroban]

Hello, sigma!

1. Jim, who is 1.8 m tall, is walking towards a lamp-post which is 3 m high.
The lamp casts a shadow behind him.
He notices that his shadow gets shorter as he moves closer to the lamp.
He is walking at 2.4 m/sec.

a) When he is 2 m from the post, how fast is the length of his shadow decreasing?

b) How fast is the tip of his shadow moving?

The answer to a) is 3.6 m/s and the answer to b) is 6 m/s but solution needed.

      A
      *
      |   *
      |       *   C
     3|           *
      |           |   *
      |        1.8|       *
      |           |           *
    --*-----------+---------------*--
      B     x     D       S       E

The lamppost is \(\displaystyle AB\,=\,3\) m.
Jim is \(\displaystyle CD\,=\,1.8\) m.

\(\displaystyle x\,=\,BD\) is his distance from the post.
\(\displaystyle S\,=\,DE\) is the length of his shadow.

Since \(\displaystyle \,\Delta CDE\,\sim\,\Delta ABE\), we have: \(\displaystyle \,\frac{S}{1.8}\,=\,\frac{x\,+\,S}{3}\;\;\Rightarrow\;\;S\,=\,\frac{3}{2}x\)

Differentiate with respect to time: \(\displaystyle \:\frac{dS}{dt}\:=\:\frac{3}{2}\left(\frac{dx}{dt}\right)\)

We are told that: \(\displaystyle \,\frac{dx}{dt}\,=\,-2.4\) m/sec.
\(\displaystyle \;\;\)(He is walking toward to the post; \(\displaystyle x\) is decreasing.)


(a) Therefore: \(\displaystyle \:\frac{dS}{dt}\:=\:\frac{3}{2}(-2.4)\:=\:-3.6\) m/sec.

      A
      *
      |   *
      |       *   C 
     3|           *
      |           |   *
      |        1.8|       *
      |           |           *
    --*-----------+---------------*--
      B     x     D     y - x     E
      : - - - - - - y - - - - - - :

The lamppost is \(\displaystyle AB\,=\,3\) m.
Jim is \(\displaystyle CD\,=\,1.8\) m.

\(\displaystyle x\,=\,BD\) is his distance from the post.
\(\displaystyle y\,=\,BE\) is the distance from the tip of his shadow to the post.
\(\displaystyle \;\;\)Then: \(\displaystyle DE\,=\,y\,-\,x\)

Since \(\displaystyle \,\Delta CDE\,\sim\,\Delta ABE\), we have: \(\displaystyle \,\frac{y}{3}\,=\,\frac{y\,-\,x}{1.8}\;\;\Rightarrow\;\;y\,=\,\frac{5}{2}x\)

Differentiate with respect to time: \(\displaystyle \:\frac{dy}{dt}\:=\:\frac{5}{2}\left(\frac{dx}{dt}\right)\)

Therefore: \(\displaystyle \:\frac{dy}{dt}\:=\:\frac{5}{2}(-2.4)\:=\:-6\) m/sec.

 

[galactus]

For question 1,

We need \(\displaystyle \L\frac{dx}{dt}\), given that \(\displaystyle \frac{dy}{dt}=-2.4\).

By similar triangles:

\(\displaystyle \L\frac{x}{1.8}=\frac{x+y}{3}\)

Now, solve for x in terms of y, differentiate and enter in your dy/dt=-2.4 and you'll have it.

For the 2nd part:

The tip of the shadow is z=x+y meters from the light, so the rate at which it is moving is given by \(\displaystyle \L\frac{dz}{dt}=\frac{dx}{dt}+\frac{dy}{dt}\)

In the first part, we found \(\displaystyle \L\frac{dx}{dt}\) when \(\displaystyle \L\frac{dy}{dt}=-2.4\), so from this find \(\displaystyle \L\frac{dz}{dt}\)

If you get hung up write back showing where you're struggling.

Soroban?. Whatcha think?.

edit: Nevermind, I arrived at the same thing only slightly different methodology.

 

[sigma]

So how come \(\displaystyle \L\frac{dx}{dt}\) is negative? Just doesn't make sense.

 

[pka]

So how come \(\displaystyle \L\frac{dx}{dt}\) is negative? Just doesn't make sense.

What does “to get shorter” mean?

 

[sigma]

for (a), it makes sense that the answer is negative now that I look closer at the question and they want a decreasing rate, but how come (b)'s answer is negative?

 

[pka]

Technically you are correct, in that ‘speed’ is a non-negative notion.
We use absolute value in connection with find speed
So perhaps (b) should not be negative.
However, from the wording one could argue either way.
But given answer suggest speed