Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am suppose to work this out somehow but I am not sure. I have tried many ways and I can't progress any further.

A streethg light his mounted at the top of a 15 ft tall pole. A man 6 ft tall walks away from the pole at a speed of 5ft/s along a straight path. How fast is the tip of the shadow moving when he is 40tf from the pole?

From this I know that I have a triangle with 2 known values on it, 15tf tall and then an opposite side of that is 6 feet tall. The rate of change of the bottom of the triangle is 5. I need to solve for c prime I am pretty sure. I am just not sure how to get that. I tried many different ways and none are correct.I tried to set up the pythagorean theorem but it didn't work and I am not sure why. I know that if I have 2 sides of a triangle I can figure out a third, so I tried to find the derivative of that but that didn't work.

share|cite|improve this question
up vote 3 down vote accepted

The $15$ foot tall lightpole and the $6$ foot tall man aren’t sides of the same triangle. You actually have two triangles. Let $A$ be the point at the base of the lightpole, $B$ the point at the top of the lightpole, $C$ the point where the man is standing, $D$ the top of his head, and $E$ the end of his shadow. Then you have a right triangle $ABE$ with a vertical side of $15$ feet and another right triangle $CDE$ with a vertical side of $6$ feet. These triangles are related in a rather special way: they’re _______?

Now let $x$ be the distance of the man from the lightpole, i.e., the length of $\overline{AC}$. You know what $dx/dt$ is. Let $y$ be the length of $\overline{AE}$, the distance from the lightpole to the tip of his shadow. You want to find $dy/dt$. If you can answer the question above, you should be able to find a relationship between $x$ and $y$ that you can use to find $dy/dt$.

share|cite|improve this answer
Okay so I don't know what it is called but the triangles have the same angels since they both are 90 degrees and then share the angle E. I don't really know how to continue I know that I could likely solve the triangles with the information provided since I know 1 angle and 3 sides and then that would tell me two the other triangle as well but I don't think that is neccessary nor do I remember how to do that. – user138246 Oct 4 '11 at 1:59
They are called similar triangles. – Altar Ego Oct 4 '11 at 2:23
@Jordan: You don’t want to solve any triangles. You need to use the fact that the two triangles are similar, so that one is a scaled-up version of the other. If you’re rusty on similar triangles, just Google on the term: there’s lots out there. – Brian M. Scott Oct 4 '11 at 2:24
I give up. This is too hard and likely won't be on a test anyways. I have other material I need to study so I can maybe not fail. – user138246 Oct 4 '11 at 3:46
@Jordan: Unfortunately, this kind of problem is bog-standard, so something like it very well could appear on an exam. Try reading Example 6 in these notes; it’s your problem with different numbers. There’s more discussion of similar triangles up in Example 4. – Brian M. Scott Oct 4 '11 at 3:55

First of all note that speed of man $(v_1)$ and speed of his shadow $(v_2)$ are constant values.Let us suppose that man starts walking from the bottom of the pole and let's observe case when man walks away another $2$ft from the point $C$ so that $CF=2$ft (see picture bellow). Note that triangle $BGD$ and triangle $BEH$ are similar so we may write next equation:

$GD : BD = EH : BE$

$2 : \sqrt{(AC)^2+(AB-CD)^2}=v_2(t_2-t_1): \sqrt{225+(v_2t_1)^2}$

where $t_1=\frac{AC}{v_1}$ ; $t_2=\frac{AC+2}{v_1}$ , so you have an equation with unknown $v_2$

enter image description here

share|cite|improve this answer
I don't follow what happened here at all. At any step basically. – user138246 Oct 4 '11 at 14:12
@Jordan, – pedja Oct 4 '11 at 16:57
I don't know what 2:square root means. – user138246 Oct 4 '11 at 17:14
@Jordan,that's means have to multiply $2\sqrt{225+(v_2t_1)^2}=v_2(t_2-t_1)\sqrt{(AC)^2+(AB-CD)^2}$ – pedja Oct 4 '11 at 17:27
Proportion of what? Where did that come from and why am I multiplying it by 2 and why is it equal to v2(t2-t1) times that thing? – user138246 Oct 4 '11 at 17:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.