What's wrong with my similar triangle for a related rate question?

In summary, the conversation is about finding the rate at which the height of a shadow decreases as a 2m tall person walks away from a lightbulb towards a wall. The problem can be solved using similar triangles, but there is confusion about the correct method. One method results in the correct answer, while another method gives a constant change rather than an accelerated one. The confusion is cleared up by considering the relationship between the constants and variables in the problem.
  • #1
student34
639
21

Homework Statement



(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.

The Attempt at a Solution



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, and it makes sense to me 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 has the wrong sign and is a constant change - not an accelerated one.

I never want to stop bashing my head into the wall until I understand why this doesn't work.
 
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  • #2
student34 said:
y/10 = 2/x When differentiated it looks like y' = -20/x^2(x') = -20/49(1.2) = -(24/49)m/s.
so y/10=2/x, how do you arrive y' = -20/x^2(x') from that?
 
  • #3
klondike said:
so y/10=2/x, how do you arrive y' = -20/x^2(x') from that?

What do you mean? This is correct isn't it?

y/10 = 2/x
y = 20/x
y = 20x^(-1)
y' = -20x^(-2)(1.2)
y' = -20(7)^(-2)(1.2)
y' = -24/49

I am quite sure that this is the right answer.
 
  • #4
student34 said:

Homework Statement



(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.


The Attempt at a Solution



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, and it makes sense to me 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 has the wrong sign and is a constant change - not an accelerated one.

I never want to stop bashing my head into the wall until I understand why this doesn't work.

What are the constants in the problem? What are the variables?

The instantaneous (momentary) x value is 7m, and therefore 10 = x + 3 at that instant. Would this relationship always hold true? What happens when x = 5m or 8m? Remember, the 10m distance is the fixed one.

Similarly, the ratio of 2/7 is true for that instant, but as the guy gets closer or further away from the bulb, the denominator will change, but the numerator remains constant (the guy does not change in height).
 
  • #5
student34 said:
What do you mean? This is correct isn't it?

y/10 = 2/x
y = 20/x
y = 20x^(-1)
y' = -20x^(-2)(1.2)
y' = -20(7)^(-2)(1.2)
y' = -24/49

I am quite sure that this is the right answer.

That part is fine.
 
  • #6
Oh, I misread the parenthese in your original post. It's correct.
student34 said:
What do you mean? This is correct isn't it?
I am quite sure that this is the right answer.
 
  • #7
student34 said:

Homework Statement



(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.


The Attempt at a Solution



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, and it makes sense to me 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 has the wrong sign and is a constant change - not an accelerated one.

I never want to stop bashing my head into the wall until I understand why this doesn't work.

Ahhhh, I see what you're saying. Is there any general rule that I am breaking here because even though I see my mistake, it is not so obvious that I won't do it again. Or, should I just stick with what is constant when forming the formula (and of course differentiate lone constants to 0)?
 
  • #8
student34 said:
Ahhhh, I see what you're saying. Is there any general rule that I am breaking here because even though I see my mistake, it is not so obvious that I won't do it again. Or, should I just stick with what is constant when forming the formula (and of course differentiate lone constants to 0)?

The aim is to derive a relationship between y and x that holds no matter how the variables change. I don't know how to explain it any more simply than that.
 
  • #9
Curious3141 said:
The aim is to derive a relationship between y and x that holds no matter how the variables change. I don't know how to explain it any more simply than that.

ok, thanks
 

Related to What's wrong with my similar triangle for a related rate question?

1. What is a similar triangle in a related rate question?

In a related rate question, a similar triangle refers to a triangle that has the same shape as another triangle, but may differ in size. This is important because similar triangles have proportional sides, which allows us to use their ratios to solve for unknown values.

2. Why is it important to have a similar triangle in a related rate question?

Having a similar triangle in a related rate question is important because it allows us to use the proportional relationships between the sides of the triangles to solve for unknown values. Without a similar triangle, it would be much more difficult to calculate the related rates.

3. How do I know if my triangle is similar in a related rate question?

To determine if two triangles are similar, we can use the Side-Angle-Side (SAS) or Angle-Angle (AA) similarity criteria. This means that either the corresponding sides of the triangles are proportional, or the corresponding angles are congruent.

4. Can I use any similar triangle in a related rate question?

No, it is important to use a triangle that is similar to the given triangle in the related rate question. Using a different similar triangle may result in incorrect solutions. It is also important to make sure that the triangles are in the same orientation.

5. What should I do if I am having trouble finding a similar triangle in a related rate question?

If you are having trouble finding a similar triangle, try using the given information to set up a proportion between the known and unknown quantities. You can also try drawing the triangles to scale to help visualize the problem and identify a possible similar triangle.

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