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Find the max angle for the resultant of two vectors

  1. Feb 10, 2009 #1
    This is a engineering question and we are not aloud to use calculus. We are told we have a boat on the bank of a river with a current to the right at 30 mph and the max speed of the boat is 25 mph. The question is, traveling at what angle gives the best chance to make it across?

    The teacher then went and broke down the idea of it and said he is looking for the angle for the 25 vector relative to the 30 vector that will give the greatest resultant angle. He also stated that when he was first prompted the question that he solved it in a rather long way using calculus but a high school teacher walked up and solved it in a very easy and simple way without calculus. Therefore, I'm assuming it has something to do with triangles but we were given no more information.

    At first I thought the answer was 135 degrees (45 degrees into the current) based off the knowledge that 45 degrees is the optimal launch angel with gravity but that is not the case here. I tested it, using sin and cos to get the resultant angel to be 55.12 degrees. To check it I then tried 37 and 53 degrees into the current and got 56.3 and 53.16. So 45 degrees into the current cannot be the correct answer. I know it has something to do with it being two vectors of the same degree rather than having something like acceleration in there. From what the teacher said I think there is something from geometry that can be used to solve this without (or maybe just a little) using sin/cos/tan. The resultant will have a greater angle as the y vector approaches equal magnitude as the x vector and will continue to rise as the y is bigger than the x vector, the resultant angle will be even greater. So I know it will be less than 45 but it can't be to low or the height will be lost.

    Any ideas on how to find the max resultant angle?
  2. jcsd
  3. Feb 10, 2009 #2


    Staff: Mentor

    So maybe you can use calculus as long as you do it silently. :wink:
  4. Feb 10, 2009 #3

    Tom Mattson

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    Let x-axis be aligned with the direction of current flow and let [itex]\theta[/itex] be the angle that the 25 mph vector makes with the postive x-axis. Now, can you write down velocity vectors for the boat relative to the water and the water relative to the Earth?
  5. Feb 11, 2009 #4
    Umm, what I'm suppose to find is the best angle to launch the boat at, using the current as the positive x axis, to find the greatest resultant angle.

    It was a extra credit question and my teacher will be going over it later today. I'll post the results he gives in case anyone else gets a question like this.
  6. Feb 11, 2009 #5
    Okay, I ended up solving it using calculus for partial credit, it was worth 5 points to the overall grade for my Engineering Mathematics class.

    I broke up the boats vector (25 mph) into x (25cos[tex]\beta[/tex] and let [tex]\beta[/tex] = 180 - launch angle or angle into the current) and y ( 25sin[tex]\beta[/tex]) directions then with the broke down vectors i created a right triangle. The base was the positive x axis (now 30-25cos[tex]\beta[/tex]) and the height is the positive y axis (25sin[tex]\beta[/tex]). The problem wants to find the launch angle best for survival so therefore the greater the angle between the current (30-25cos[tex]\beta[/tex]) and the hypotenuse (now resultant) is the key, I will call this [tex]\theta[/tex]. This angle needs to be maximized to be as close to 90 as possible to get closest to a straight line across the body of water. To find the angle I used tan[tex]\theta[/tex]=(25sin[tex]\beta[/tex])/(30-25cos[tex]\beta[/tex]) and then turned into [tex]\theta[/tex]= arctan ((25sin[tex]\beta[/tex])/(30-25cos[tex]\beta[/tex]). To maximize [tex]\theta[/tex] you take the derivative and it gets really messy so I graphed the [tex]\theta[/tex]= equation and in the graph [tex]\theta[/tex]=y and [tex]\beta[/tex]=x and had it calculate the max. The answer was [tex]\beta[/tex]=33.557, the launch angle relative to the current is 146.443, and the resultant angle [tex]\theta[/tex] = 56.443

    The teacher showed that the problem could also be view as a rhombus. Using the 30 as the base, 25 as the sides and the resultant is the diagonal from bottom left to top right. Then ignore the top left side of the diagonal, as you notice the 25 vector when moved the resultant also moves to meet it. So therefore the 25 vector side can be viewed as the radius of a circle. From geometry, we can conclude that the line with the greatest angle then is tangent to the circle. From there we use the tangent to know that it and the 25 vector form a right angle and then we have the length of the two sides so we can use tangent to find [tex]\theta[/tex] the angle between the 30 and the hypotenuse.

    This might be a good problem for teachers elsewhere to show the importance of geometry and other maths and how they can make real world problems easy. I know my classes in high school always moaned and groaned about how they're were never gonna need math, they still may not but at least they will know more about how its applied.

    Edit: Sorry for the typo's earlier, I scrolled up and read I had some angel's instead of angle's. Would an admin be able to change the tags and add "math applied","geometry applied", "tangent". Thanks
    Last edited: Feb 11, 2009
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