What is wrong with my similar triangle method to finding a related rate?

[The Student]

*This time I am sure that I am doing something wrong.

(I don't know how to draw a picture here so I will just explain the dimensions of the similar triangles)

A lightbulb on a floor shines at a 2m tall guy who is walking away from it towards a wall. As he gets closer to the wall, his shadow gets smaller and smaller on the wall. The lightbulb is 10m from the bottom of the wall. This makes a triangle where the base is 10m and the height of it is the height of the shadow (y). At what rate is the height of the shadow decreasing if the guy walks at a rate of 1.2m/s and is 7m (we will call x) from the lightbulb. This information creates a similar small triangle with a base of x = 7m (his distance from the lightbulb) and a height (his height) of 2m.

So, in my notes my instructor found the related rate by using a similar triangle method of y/10 = 2/x When differentiated it looks like y' = -20/x^2(x') = -20/49(1.2) = -(24/49)m/s. I was fine with this until I tried a different similar triangle method. I tried putting y/(x+3) = 2/7 which gives me the correct x and y dimensions of the triangles. But when I differentiate the equation, I don't get the right answer. Instead, I get:
y = (2x)/7 + 3/7
y' = (2(x'))/7 + 0
y' = (2(1.2))/7
y' = 12/35 which also has the wrong sign.

 

[soroban]

Hello, The Student!

A lightbulb on a floor shines at a 2m tall man who is walking away from it towards a wall.
s he gets closer to the wall, his shadow gets smaller and smaller on the wall.
The lightbulb is 10m from the bottom of the wall.
This makes a triangle where the base is 10m and the height of it is the height of the shadow (y).
At what rate is the height of the shadow decreasing if the guy walks at a rate of 1.2m/s and is 7m from the lightbulb?

This information creates a similar small triangle with a base of x = 7m (his distance from the lightbulb)
and a height (his height) of 2m. . This is a bad idea!
So, in my notes my instructor found the related rate by using a similar triangle method of y/10 = 2/x
When differentiated it looks like y' = -20/x^2(x') = -20/49(1.2) = -(24/49)m/s.

I was fine with this until I tried a different similar triangle method.
I tried putting y/(x+3) = 2/7 which gives me the correct x and y dimensions of the triangles.

You are inserting constant values too early.

If you insist that \(\displaystyle x = 7\), you have a static situtation . . . no movement.

                          *
                      *   |
                  *       |y
              *   |2      |
          *       |       |
      *-----------*-------*
            7         3

Then we have: .\(\displaystyle \frac{y}{10} \:=\:\frac{2}{7} \quad\Rightarrow\quad y \:=\:\frac{20}{7}\)

We know the height of the shadow,
. . but nothing about the rate of change of the shadow.



Advice: don't use \(\displaystyle x = 7\) until the very end.

                        *
                    *   |
                *       |y
            *   |2      |
        *       |       |
    *-----------*-------*
    : - - x - - :
    : - - - - 10  - - - :

Now we have: .\(\displaystyle \dfrac{y}{10} \:=\: \dfrac{2}{x} \quad\Rightarrow\quad y \:=\:20x^{-1}\)

Differentiate with respect to time: .\(\displaystyle \dfrac{dy}{dt} \:=\:-20x^{-2}\dfrac{dx}{dt}\)


Now substitute the known values: .\(\displaystyle \frac{dx}{dt} = 1.2,\;x = 7\)

.\(\displaystyle \dfrac{dy}{dx} \:=\:-\dfrac{20}{7^2}(1.2) \:=\:-\dfrac{24}{49}\text{ m/sec}\)

I was fine with this until I tried a different similar triangle method.
I tried putting y/(x+3) = 2/7 which gives me the correct x and y dimensions of the triangles.

It's obvious that you already insisted that \(\displaystyle x = 7.\)

 

[The Student]

Hello, The Student!


You are inserting constant values too early.

If you insist that \(\displaystyle x = 7\), you have a static situtation . . . no movement.

                          *
                      *   |
                  *       |y
              *   |2      |
          *       |       |
      *-----------*-------*
            7         3

Then we have: .\(\displaystyle \frac{y}{10} \:=\:\frac{2}{7} \quad\Rightarrow\quad y \:=\:\frac{20}{7}\)

We know the height of the shadow,
. . but nothing about the rate of change of the shadow.



Advice: don't use \(\displaystyle x = 7\) until the very end.

                        *
                    *   |
                *       |y
            *   |2      |
        *       |       |
    *-----------*-------*
    : - - x - - :
    : - - - - 10  - - - :

Now we have: .\(\displaystyle \dfrac{y}{10} \:=\: \dfrac{2}{x} \quad\Rightarrow\quad y \:=\:20x^{-1}\)

Differentiate with respect to time: .\(\displaystyle \dfrac{dy}{dt} \:=\:-20x^{-2}\dfrac{dx}{dt}\)


Now substitute the known values: .\(\displaystyle \frac{dx}{dt} = 1.2,\;x = 7\)

.\(\displaystyle \dfrac{dy}{dx} \:=\:-\dfrac{20}{7^2}(1.2) \:=\:-\dfrac{24}{49}\text{ m/sec}\)



It's obvious that you already insisted that \(\displaystyle x = 7.\)

Ok, I see what I did wrong now - thank-you so much.