Finding the minimum length of a median in a triangle

In summary, In right triangle ABC, AD is a median to BC. AB+BC=4. A) Mark BD with an x, and express the lengths of the adjacent and opposite through x. B) Find the x for which the length AD is minimal.
  • #1
Femme_physics
Gold Member
2,550
1

Homework Statement



In a right triangle ABC, AD is a Median to BC (see drawing)
Given:
AB+BC = 4

A) Mark BD with an x, and express the lengths of the adjacent and opposite through x.
B) Find the x for which the length AD is minimal

http://img641.imageshack.us/img641/6798/abctriangle.jpg [Broken]

The Attempt at a Solution



Solving A was pretty easy

A)
http://img405.imageshack.us/img405/1696/47912199.jpg [Broken]Solving B is far more challenging, I first tried to express AB as a function of x. This is what I got:

http://img33.imageshack.us/img33/5889/26052312.jpg [Broken]

But if I try to take its derivative I don't get a function, I get a value of 7. I must've done something wrong along the way, right?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Your application of the Square root is incorrect.

Think about this:

SQRT(4+4)

That does not equal 2+2
 
  • #3
Hmmm...

And what if I multiply them?

Such as,

(4-2x) * x?
 
  • #4
sqrt(ab) = sqrt(a)*sqrt(b)

(Presuming you mean to be solving for AD in your last image...)
However, the squareroot is probably irrelevant for what you're trying to solve. The max and min values for AD^2 will be at the same x as AD (when AD is positive, which is all you're interested in anyhow since we're looking at a length). So you can just solve the derivative inside the sqrt and the values will be the same in the range you want.
 
  • #5
The max and min values for AD^2 will be at the same x as AD

Really? That's an interesting fact. The derivative of something under a square root is always the same derivative as the thing with the square root? (talking only about positive values).

So you can just solve the derivative inside the sqrt and the values will be the same in the range you want.

That would be
(4-2)+1
The derivative would be equal 3, without any unknowns!

Does it make sense?
 
  • #6
You still have the squares, I'd expand the square of AB and then add your x^2. Should be a polynomial, then take the derivative, which should be a linear function (not a constant) and solve for x. (double check my work, but you'll be taking the derivative of something like 16-16x+5x^2 - which is what the 'under the sqrt' expression should come out to be when expanded)
 
  • #7
Wait, I think I'd like to first of all formally acknowledge my typo of AB instead of AD. You're right.

I'd expand the square of AB and then add your x^2. Should be a polynomial, then take the derivative, which should be a linear function (not a constant) and solve for x.

You mean expand this? (did you write AB mistakenly as well?)

[tex]AD = x^{2} + (4-2x)^{2}[/tex]
 
  • #8
Femme_physics said:
You mean expand this? (did you write AB mistakenly as well?)

[tex]AD = x^{2} + (4-2x)^{2}[/tex]

Yes! :smile:


Actually, what you have is:

[tex]AD^2 = x^{2} + (4-2x)^{2}[/tex]

The point made before, is that if AD takes on a minimum value, AD2 will also take on a minimum value.
And it's easier to take a derivative from a formula that does not contain a square root.
 
  • #9
Yea, I wrote AB originally, and corrected it within a few minutes of posting originally. No biggie ;)
 
  • #10
The point made before, is that if AD takes on a minimum value, AD2 will also take on a minimum value.
And it's easier to take a derivative from a formula that does not contain a square root.

So if I take the derivative of [tex]AD = x^{2}+(4-2x)^{2}[/tex]
I'll get the same minimum value if I take the derivative of this:


[tex]AD = \sqrt{x^{2}+(4-2x)^{2}}[/tex]

In that case, it's an interesting fact... and curious that I can't seem to grasp intuitively how come they're the same.

But okay, I'll take the derivative of
[tex]AD = x^{2}+(4-2x)^{2}[/tex]

So I get that the derivative of AD is

[tex]AD = 2x-4x[/tex]


and x=0
Makes AD 0. Unless I took the derivative wrongly again?
 
Last edited:
  • #11
Femme_physics said:
So if I take the derivative of [tex]AD^2 = x^{2}+(4-2x)^{2}[/tex]
I'll get the same minimum value if I take the derivative of this:

[tex]AD = \sqrt{x^{2}+(4-2x)^{2}}[/tex]

Well, you will not get the same minimum value, but they both get their minimum value at the same x value.

Femme_physics said:
In that case, it's an interesting fact... and curious that I can't seem to grasp intuitively how come they're the same.

I'll try to explain.
If AD takes its minimum somewhere, than the value of AD is slightly bigger for smaller and larger x.
Since AD2 = AD x AD, that means that AD2 will also have a value that is slightly bigger for smaller and larger x.

Femme_physics said:
But okay, I'll take the derivative of
[tex]AD = x^{2}+(4-2x)^{2}[/tex]

So I get that the derivative of AD is

[tex]AD = 2x-4x[/tex]

and x=0
Makes AD 0. Unless I took the derivative wrongly again?

Sorry, the derivative is not right.
Could you first expand the expression, before taking the derivative?
 
  • #12
I can, but can you tell me how do I make a "new line" with that Online LaTeX editor you've linked me to? Pressing "enter" doesn't cut it.
 
  • #13
AD' and (AD^2)' will be zero at the same x-coordinate. They are NOT the same derivative function, but for a min/max value problem they can provide the same information.

Do you have access to a graphing calculator or graphing software? Maybe it will help visualize what the two functions (AD and AD^2) are doing.
 
  • #14
Femme_physics said:
I can, but can you tell me how do I make a "new line" with that Online LaTeX editor you've linked me to? Pressing "enter" doesn't cut it.

Uhh... yeah... sorry... that is a bit tricky.

I'd suggest to use [ tex ] tags repeatedly for each separate line.


Oh, and for the explanation of how the minimum values work, check this out: :smile:
http://www.wolframalpha.com/input/?i=plot+x^2+(4-2x)^2,+sqrt(x^2+(4-2x)^2)

Maybe it helps to visualize it.

@mege: good suggestion :)
 
  • #15
Femme_physics said:
I can, but can you tell me how do I make a "new line" with that Online LaTeX editor you've linked me to? Pressing "enter" doesn't cut it.

erm, yea, do what she said - just use separate tex brackets... i can't figure out how to get a new line to work using the doc.
 
  • #16
Last edited by a moderator:
  • #17
(LaTeX is stupid, let me me test it first)

[tex]AD = x^{2}+(4-2x)^{2} // AD = x^{2}+(4-2x)+(4-2x) // AD = 2x+(-2)+(-2) // AD = 2x-4[/tex]Edit: Argh! Didn't work!PS (posted graph in my previous reply :) )
 
  • #18
Is there a LaTeX button on the top bar like there is for "quote"? It would make my life a lot easier when editing for new lines...
 
  • #19
Another reason this works is think about what the derivative of sqrt of some function f(x) is (using chain rule):

[tex]
\frac{d}{dx} \sqrt{f(x)} = \frac{1}{2}\frac{1}{\sqrt{f(x)}} f'(x)
[/tex]

Whenever f'(x)=0 (which will end up being the numerator of the fraction) the entire derivative is zero. We are just skipping the step of writing what will end up being a rather complex looking fraction by just evaluating the meat of the expression.
 
  • #20
mege said:
erm, yea, do what she said - just use separate tex brackets... i can't figure out how to get a new line to work using the doc.

@mege: If you really must know, you need to use a latex "environment", that is, to put it for instance like this:

[ tex ]
\begin{matrix}
a \\
b \\
c
\end{matrix}
[ /tex ]
 
Last edited:
  • #21
Femme_physics said:
(LaTeX is stupid, let me me test it first)

[tex]AD = x^{2}+(4-2x)^{2} // \underline{AD = x^{2}+(4-2x)+(4-2x)} // AD = 2x+(-2)+(-2) // AD = 2x-4[/tex]


Edit: Argh! Didn't work!


PS (posted graph in my previous reply :) )

The underlined portion (and what comes after it) is incorrect.

[tex]
(a-b)^2 = (a-b) * (a-b) = a^2 - 2ab + b^2
[/tex]
 
  • #22
Femme_physics said:
Is there a LaTeX button on the top bar like there is for "quote"? It would make my life a lot easier when editing for new lines...

I don't understand. What do you mean?

I suggest you enter each line in the LaTeX editor separately.
Then copy and paste it in the reply box and put [ tex ] tags around it.

Btw, nice graph! :smile:
 
  • #23
I don't understand. What do you mean?

I suggest you enter each line in the LaTeX editor separately.
Then copy and paste it in the reply box and put [ tex ] tags around it.

I'm sorry, LaTeX sucks! It's also buggy (I posted example here -> https://www.physicsforums.com/showthread.php?t=498142 )

And it takes some work to get new lines, too. Does this thing really worth the pain, I ask myself?

Let me just go back to Microsoft Equation Screenshots, please.
mege said:
The underlined portion (and what comes after it) is incorrect.

[tex]
(a-b)^2 = (a-b) * (a-b) = a^2 - 2ab + b^2
[/tex]

My bad. *slaps forehead*

Corrected:

http://img204.imageshack.us/img204/2159/adadz.jpg [Broken]Looks good so far?

Btw, nice graph!

Thanks! ^^
Another reason this works is think about what the derivative of sqrt of some function f(x) is (using chain rule):

I think it makes more sense to me if I think about parabola's than if I think about the chain rule. Because I know that parabolas become wider or thinner depends on the coefficient of the unknown that's raised to the power of 2.
 
Last edited by a moderator:
  • #24
d/dx 4x^2 = 8x (not 16x)


(I got a rational fraction for a result in the end)
 
  • #25
Femme_physics said:
I'm sorry, LaTeX sucks! It's also buggy (I posted example here -> https://www.physicsforums.com/showthread.php?t=498142 )

And it takes some work to get new lines, too. Does this thing really worth the pain, I ask myself?

Yeah, LaTeX on PF has an issue with refreshing that's been bugging *everyone* on PF.
The workaround is the hit the refresh button on your web browser.

Good call that you posted it in the forum feedback.
Perhaps a PF Admin will do something about it now. :)

Femme_physics said:
Let me just go back to Microsoft Equation Screenshots, please.

Whatever pleases you! :)

Femme_physics said:
Looks good so far?

Your derivative of 4x2 is not right...
 
  • #26
Fixed, I believe :)

http://img51.imageshack.us/img51/3682/corrected.jpg [Broken]


Yeah, LaTeX on PF has an issue with refreshing that's been bugging *everyone* on PF.
The workaround is the hit the refresh button on your web browser.

Good call that you posted it in the forum feedback.
Perhaps a PF Admin will do something about it now. :)

If it hasn't been solved so far and everybody knows about it, I won't be holding my breath :/



-Or
 
Last edited by a moderator:
  • #27
Femme_physics said:
Fixed, I believe :)

You can have my seal if you want it! :smile:
 
  • #28
Femme_physics said:
Fixed, I believe :)





If it hasn't been solved so far and everybody knows about it, I won't be holding my breath :/



-Or


/clap

A few quick tests can verify this.

I checked values within .1 of the length found (plus this gives me an opportunity to further play with latex line breaks...)

[tex]
\begin{flalign*}
\overline{AC} & = & & \sqrt{2.4^2+1.6^2} & = & 2.88 \\
& = & & \sqrt{2.5^2+1.5^2} & = & 2.91 \\
& = & & \sqrt{2.6^2+1.4^2} & = & 2.95 \\

\end{flalign*}
[/tex]
 
Last edited by a moderator:
  • #29
[[Sorry it took me so long to reply -- I lost internet connection for a while ]]


Excellent! Now that I have the value for the minimum (and I did check to make sure it's a minimum), I plug this value of 1.6 into the original AD function and I get that

x = 1.7888

So, I do believe I have my final answer here, right?

Edit Realised you can't solve it this way, sorry will come back.
Yes, reading your post I was like that... :bugeye:

But thanks for your feedback. As I've said before, you and ILS in the same thread is a total... ;)


/clap

A few quick tests can verify this.

I checked values within .1 of the length found (plus this gives me an opportunity to further play with latex line breaks...)

Is that the equation for AD you put in those tests? Because I seem to have a slightly different equation and answer to 1.6

http://img690.imageshack.us/img690/6767/adadadadad.jpg [Broken]
 
Last edited by a moderator:
  • #30
Erm, yea - That's going to be AC... my bad - the comparisons should still end up being similar since the outside hypotenuse should scale the same. I'm glad you're catching my tangents to the problem.

x is x is 1.6 regardless. 1.7888 is length of AD at that value. So 1.6 is your answer ("Find the x for which the length AD is minimal")
 
  • #31
Oh, you're correct! I forgot the actual question lol. :) Yes, I was solving for x, not for AD. Many thanks mege and ILS! You two are brilliant!
 
  • #32
Edit Realised you can't solve it this way, sorry will come back.

Yes, reading your post I was like that...

But thanks for your feedback. As I've said before, you and ILS in the same thread is a total... ;)

So here is the amended version using implicit differentiation.

Let y = AD2

[tex]\begin{array}{l}
y = {x^2} + {\left( {4 - 2x} \right)^2} \\
y = {x^2} + 4{x^2} - 16x + 16 \\
y = 5{x^2} - 16x + 16 \\
dy = \left( {10x - 16} \right)dx \\
\frac{{dy}}{{dx}} = \left( {10x - 16} \right) = 0 \\
x = 1.6 \\
\end{array}[/tex]

Implicit differentiation is a good trick to learn as it can significantly reduce the work.

go well
 

1. What is the definition of a median in a triangle?

A median in a triangle is a line segment that connects a vertex of the triangle to the midpoint of the opposite side.

2. How many medians does a triangle have?

A triangle has three medians, one from each vertex.

3. How do you find the minimum length of a median in a triangle?

The minimum length of a median in a triangle can be found by using the formula: minimum median length = 1/2 * (side length opposite the median)

4. Can the minimum length of a median in a triangle be zero?

No, the minimum length of a median in a triangle cannot be zero. This is because the side opposite the median must have a positive length, and the midpoint of that side must also have a positive length.

5. Is the minimum length of a median in a triangle always the same?

No, the minimum length of a median in a triangle can vary depending on the length of the side opposite the median. The longer the side, the longer the minimum length of the median will be.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Replies
1
Views
870
  • Precalculus Mathematics Homework Help
Replies
27
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
Replies
1
Views
729
  • Precalculus Mathematics Homework Help
Replies
18
Views
460
Replies
2
Views
1K
Replies
1
Views
1K
Back
Top