Optimization Using Trigonometry Instead of Calculus

P

parallax7

New member
Joined
Nov 3, 2019
Messages
10
A person in a boat 4km off the coast. The person needs to reach a point 10km down and along the coast in the least possible time. The person can row 6km/hr and run 8km/hr. How far down the coast should he land the boat.

I know how to solve this using calculus. I create a function of time, take the derivative and set it equal to zero, and then solve for x. Then I check the critical points to find the minimum. However, my professor wants me to try and solve it only using trigonometry and not calculus. I tried a could of things, such as creating a function for time, where the angle theta is the input variable, but after that I was just plugging in values to try and find the optimal angle theta and then eventually solving for x.

Thank you
 
Please draw a diagram. Also solve this with calculus so you know the answer and then try solving it with trigonometry. Hopefully the function you want to minimize might be a quadratic equation which you can do with out calculus?
 
parallax7 said:
A person in a boat 4km off the coast. The person needs to reach a point 10km down and along the coast in the least possible time. The person can row 6km/hr and run 8km/hr. How far down the coast should he land the boat.

I know how to solve this using calculus. I create a function of time, take the derivative and set it equal to zero, and then solve for x. Then I check the critical points to find the minimum. However, my professor wants me to try and solve it only using trigonometry and not calculus. I tried a could of things, such as creating a function for time, where the angle theta is the input variable, but after that I was just plugging in values to try and find the optimal angle theta and then eventually solving for x.Thank you
Click to expand...
Can you write the distance function - that you minimized through calculus - using trigonometric functions?
 
I attached two images. That's how I solved it using calculus. He just told me to try and solve it only using trigonometry and then present it to the class for extra credit. I have no idea how to solve it only using trigonometry. I feel like there are too many unknowns. One method I tried was just just writing a function for time, using theta as my input variable and then I just kept plugging in values to find the optimal angle theta. Eventually I did converge on the same answer that I received using the calculus. But I suspect that there might be a better way to solve it.
 

Attachments

  • 20191103_184815.jpg
    20191103_184815.jpg
    1.3 MB · Views: 7
  • 20191103_184752.jpg
    20191103_184752.jpg
    1.4 MB · Views: 7
You seem to be all over the place. In one part you had a piece of the derivative being \(\displaystyle \frac{1}{8(10-x)}\) and in another part you had \(\displaystyle \frac{10-x}{8}\). Well which is it? Can you write the formula that you want to minimize using trigonometry?
 
Jomo said:
You seem to be all over the place. In one part you had a piece of the derivative being \(\displaystyle \frac{1}{8(10-x)}\) and in another part you had \(\displaystyle \frac{10-x}{8}\). Well which is it? Can you write the formula that you want to minimize using trigonometry?
Click to expand...

No. The original function for time is
t = sqrt(16+x^2)/6 + (10-x)/8

The derivative
t' = 2x/(12sqrt(16+x^2) - 1/(8(10-x))

The rest was just seeing if I could simplify it more. Then I set t' = 0 and then solved for x to find a critical point. Then I found the minimum. So when did I ever write t' is both 1/(8(10-x)) and (10-x)/8?
 
Last edited:
If you guys don't have any ideas for me to try that's fine. I have a 100%, but I was still hoping to figure it out. I couldn't find an inkling of anyone who solved a similar problem using only trigonometry without any calculus.
 
parallax7 said:
No. The original function for time is
t = sqrt(16+x^2)/6 + (10-x)/8

The derivative
t' = 2x/(12sqrt(16+x^2) - 1/(8(10-x))

The rest was just seeing if I could simplify it more. Then I set t' = 0 and then solved for x to find a critical point. Then I found the minimum. So when did I ever write t' is both 1/(8(10-x)) and (10-x)/8?
Click to expand...
Since the derivative of 10-x is -1 we have derivative of \(\displaystyle \frac{10-x}{8}\) = -1/8 Not - 1/(8(10-x))

Now go back to your drawing, label an angle and try to get t = sqrt(16+x^2)/6 + (10-x)/8 in terms of that angle.
 
Jomo said:
Since the derivative of 10-x is -1 we have derivative of \(\displaystyle \frac{10-x}{8}\) = -1/8 Not - 1/(8(10-x))

Now go back to your drawing, label an angle and try to get t = sqrt(16+x^2)/6 + (10-x)/8 in terms of that angle.
Click to expand...
Okay. You're right. My apologies. One of the papers that I posted had a mistake on it. Thanks for the help. How do I minimize the function for t after rewriting it in terms of theta without using calculus though? As long as I know how to do that I think I can figure out the rest.
 
Yeah. Sorry about that. I have a lot of papers here and I posted one with a mistake in it.
 
In the beginning, you formulated t = f(x)

Work with that.

Show that it is an equation of a parabola - and locate its vertex for minimum value.
 
Here are the images that I should have posted. I'll try and work on the trigonometry now. Thank you.
 

Attachments

  • 20191105_072344.jpg
    20191105_072344.jpg
    2.9 MB · Views: 5
  • 20191105_072326.jpg
    20191105_072326.jpg
    1.4 MB · Views: 5
Subhotosh Khan said:
In the beginning, you formulated t = f(x)

Work with that.

Show that it is an equation of a parabola - and locate its vertex for minimum value.
Click to expand...


Correct me if I did this wrong.
t(θ)=(4csc(θ))/6 + (10-4cot(θ))/8

I checked the minimum of an upper and lower bound and the minimum was 4.54. Which matches the minimum that I received using the calculus, which was (12sqrt(7))/7. The y value also matches and is 1.69. The angle θ I believe is 41.4 degrees.

I'm not really sure how to find the vertex of this without using my calculator though.
 
Last edited:
parallax7 said:
Correct me if I did this wrong.
t(θ)=(4csc(θ))/6 + (10-4cot(θ))/8

I checked the minimum of an upper and lower bound and the minimum was 4.54. Which matches the minimum that I received using the calculus, which was (12sqrt(7))/7. The y value also matches and is 1.69. The angle θ I believe is 41.4 degrees.

I'm not really sure how to find the vertex of this without using my calculator though.
Click to expand...
You'll need to make it a polynomial of cot(θ) first - then convert it to a quadratic and then find the vertex. Kind of tedious.....
 
parallax7 said:
Okay. You're right. My apologies. One of the papers that I posted had a mistake on it. Thanks for the help. How do I minimize the function for t after rewriting it in terms of theta without using calculus though? As long as I know how to do that I think I can figure out the rest.
Click to expand...
Before you can try to minimize a function you first need the function. Maybe after looking at this function it may be obvious how to minimize, but you need to have this function before you can do anything.
 
parallax7 said:
Here are the images that I should have posted. I'll try and work on the trigonometry now. Thank you.
Click to expand...
The numbers would have been much smaller (less of a chance for a mistake) if you had divided both sides by 2 at the very beginning. Your work is fine!
 
You must log in or register to reply here.
Top