Finding the Minimum Time

  • #1
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Here's the problem:

https://i.stack.imgur.com/rUVvu.png
rUVvu.png


What I did:

[2]: https://i.stack.imgur.com/AX2Ye.png
AX2Ye.png

[3]: https://i.stack.imgur.com/K1Zbi.png
K1Zbi.png

To those who could not understand what I read: So basically I used some geometry to to get $l_1$ and $l_2$ in terms of $r$ and the assumed angle $y$. Now, I assumed that if the ant can climb straight up at a speed $v$, then it should be able to climb up the $l_2$ part with a speed $v \sin{y}$ and similarly for the downward case. Using $t = \frac{l}{v}$ for constant speed, I got $$\boxed{t(y) = \frac{3r}{v\sqrt{2}} \cdot \frac{1}{\sin{y}}}$$. This should be minimised, but minimising this function is not giving me the correct answer. Can someone help me please?

And please post the correct solution you are getting.
 

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  • #2
BvU
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I assumed that if the ant can climb straight up at a speed $$v$$, then it should be able to climb up the $$l_2$$ part with a speed $$v \sin{y}$$
My reading of the exercise statement is that up at any angle goes at ##v## and down at any angle goes at ##2v##. Apparently you already have the correct answer, so you can check.
 
  • #3
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But even then the answer is not matching with the correct answer.
Then the answer would be ##t = \frac{l_1}{2v} + \frac{l_2}{v} = \frac{r\sqrt{2}}{2v} \cdot (3-\cos{2y})## This should give the minima at ##\cos{2y} = 1## or ##y = 0##. At this angle the answer should be ##\boxed{t = \frac{r}{v\sqrt{2}}}##, but that is not the correct answer.
 
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  • #4
ehild
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To those who could not understand what I read: So basically I used some geometry to to get $l_1$ and $l_2$ in terms of $r$ and the assumed angle $y$. Now, I assumed that if the ant can climb straight up at a speed $v$, then it should be able to climb up the $l_2$ part with a speed $v \sin{y}$ and similarly for the downward case. Using $t = \frac{l}{v}$ for constant speed, I got $$\boxed{t(y) = \frac{3r}{v\sqrt{2}} \cdot \frac{1}{\sin{y}}}$$. This should be minimised, but minimising this function is not giving me the correct answer. Can someone help me please?
The problem says that the ant can climb the upward wire segment at speed v, and the downward wire segment with speed 2v. It does not mean vertically upward and downward!
You should type in your solution, or at least, provide a clear picture.
 
  • #5
BvU
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but that is not the correct answer
Obviously not: ##2r\over 2v## is the absolute minimum, so how you can come out lower is a riddle to me -- after all you haven't posted what ##y## is -- or should I decipher that from the photo ?

I also don't see the 38 cm come back in your answer, which is very strange.
 
  • #6
ehild
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I also don't see the 38 cm come back in your answer, which is very strange.
The 38 cm does not need to "come back": It means that the wire segments are too short to make a straight path over the groove. The boy connects two wires at right angles and put it on the groove, one segment making a certain angle θ with the horizontal. So the path for the ant consists of two line segments, the legs a and b of a right triangle, of hypotenuse 2r, and angle θ. The time can be written as function of the angle θ, and the minimum should be found in terms of θ.

upload_2019-2-18_12-48-52.png
 

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  • #7
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If it does not mean going exactly straight down with ##2v## and straight up with ##v##, do you think it means going anywhere making an angle ##0 \leq \theta \leq 180## means going upward with ##v## and going anywhere making an angle ##180 \leq \theta \leq 360## means going downward with ##2v##? If that is the case, this should be my solution then:

$$t = \frac{l_1}{2v} + \frac{l_2}{v}$$ Obviously by cosine rule in the triangle ##\triangle{AOB}## and ##\triangle{BOD}## we get $${l_1}= r \sqrt{2} (1-\cos{2y})$$ and $$l_2 = r \sqrt{2} (1+\cos{2y})$$ by using the fact that ##x+y = 90^{\circ}##. Inserting the values of ##l_1## and ##l_2## in the first equation we get $$\boxed{t = \frac{r}{v \sqrt{2}}}$$ This is not, however, the correct answer.

Here is the image for reference: https://i.stack.imgur.com/JZFlu.png
The green point denotes the center of the semi-circle.
 
  • #8
ehild
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If it does not mean going exactly straight down with ##2v## and straight up with ##v##, do you think it means going anywhere making an angle ##0 \leq \theta \leq 180## means going upward with ##v## and going anywhere making an angle ##180 \leq \theta \leq 360## means going downward with ##2v##? If that is the case, this should be my solution then:

$$t = \frac{l_1}{2v} + \frac{l_2}{v}$$ Obviously by cosine rule in the triangle ##\triangle{AOB}## and ##\triangle{BOD}## we get $${l_1}= r \sqrt{2} (1-\cos{2y})$$ and $$l_2 = r \sqrt{2} (1+\cos{2y})$$ by using the fact that ##x+y = 90^{\circ}##. Inserting the values of ##l_1## and ##l_2## in the first equation we get $$\boxed{t = \frac{r}{v \sqrt{2}}}$$ This is not, however, the correct answer.

Here is the image for reference: https://i.stack.imgur.com/JZFlu.png
The green point denotes the center of the semi-circle.
Why do you use cosine rule? You have got a single right triangle, with hypotenuse 2r, and the length of the legs can be expressed wit the sine and cosine of a single angle, y for example.
Check the equations in red. .I think, you have errors.
 
  • #9
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^True. That is a mistake. Continuing on what you mentioned, we find ##l_1, \text{and}, l_2##. This gives:

$$t = \frac{r}{v} \cdot (\cos{y} +2\sin{y})$$. But I believe using trigonometry to set ##y## in terms of knowns should give the answer, since ##y## is not a variable. Anyone help me with this trigonometry?
 
  • #10
BvU
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Google is your friend. Do you know how to solve e.g. ##a\sin\alpha + b \sin\alpha = 0 ## ?
PS what do you want to do with this expression ?
 
  • #11
ehild
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^True. That is a mistake. Continuing on what you mentioned, we find ##l_1, \text{and}, l_2##. This gives:

$$t = \frac{r}{v} \cdot (\cos{y} +2\sin{y})$$. But I believe using trigonometry to set ##y## in terms of knowns should give the answer, since ##y## is not a variable. Anyone help me with this trigonometry?
Why not is y a variable? What should it be in order to make t minimum?
(Here you might use that the wires are 38 cm long.)
 
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  • #12
BvU
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If I were a lazy ant, I'd want to go downwards as much as possible. How much is that ?
 
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  • #13
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I think the boy has made the setup, fixed it, and now the ant will travel there. There is no such thing as changing ##y## now.
Google is your friend. Do you know how to solve e.g. asinα+bsinα=0asin⁡α+bsin⁡α=0a\sin\alpha + b \sin\alpha = 0 ?
PS what do you want to do with this expression ?
I tried solving it, and I know how to. But I'm one equation less, and thus the equation keeps giving me ##1=1## back.
 
  • #14
haruspex
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I think the boy has made the setup, fixed it, and now the ant will travel there. There is no such thing as changing ##y## now.
No, you are misreading the question (but I agree it is unclear). There is no way to solve it under your interpretation. The intended question is: of all the ways the boy might position the wire, which allows the ant to cross in the least time.
 
  • #15
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What's the longest downslope you can make with a 38 cm wire and a 40 cm gutter ?
 

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