Question

Find the point on the curve y=cos(x) closest to the point (1,1).

Answer 1

You have two points. A point (x,y) that is on the curve y = cos(x) and a point (1,1)
You want (x,y) to be as close as possible to (1,1) In other words. You want the distance between the two points to be a minimum. Minimum! What do we know about minimums? When dy/dx = 0!
So first the equation.
Distance formula: \[\sqrt{(x-1)^2 + (y-1)^2}\]
Now sub in cos x for y
\[\sqrt{(x-1)^2 + (\cos x-1)^2}\]
Derive this, and set the derivative = to 0. The resulting x value is where the distance is a minimum. Plug in the x into the original equation to find y.

Answer 2

derivative for that is ridiculous, i have no clue where to go with that thing lol

Answer 3

The distance of the two points dont necessarily approach 0. The point may not even be on the graph. As in this case it is not, cos(1) = not 1

You have an equation for the Distance D correct? I put that above. You want this distance to be as small as possible, not necessarily 0 (it might in some case). So you want to find the minimum values. You could graph it, and try to visually look at the min (or using a graphing calculator function). But we know, that when the derivative of a function is 0, the function has extrema, in this case a min.

So we derive the Distance function, and where the derivative = 0, we have our min,.

Answer 4

@Jeremy0203 which course does this problem come from?

Answer 5

calc I

Answer 6

My textbook that covers doing this type of equation is like half a page and doesn't explain in a good way at all how this is done

Answer 7

Its calc1 yeah. Its optimization. Its abuot problems where you want to find a max or min, with regard to area, distance, volume, prices, things like that.
Here this might help
http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx

Answer 8

@NobodyOwens that is the max/min for area, not distance!!

Answer 9

No thats distance too. Here we did the same thing, but we derived the equation for distance. It can be applied to many different things.
If you want a distance specific example
http://tutorial.math.lamar.edu/Classes/CalcI/MoreOptimization.aspx
See Example 3

Answer 10

It all depends on the problem, but in the end we are deriving to determine max/mins

Answer 11

it seems this problem is over my head!!! http://www.assignmentexpert.com/free-questions/question-on-math-calculus-4905.html

Answer 12

I got (0.79,0.70)

Answer 13

I substituted in cos(x) for y in the distance formula. Then I negated the sq rt since minimizing the radical minimizes the whole equation and factored it out and solved for x which was 0.79. Then i put that back into the original equation to solved y this giving me the second point.

Answer 14

Sound bout right?

Answer 15

@NobodyOwens

Answer 16

Yup! And looks right graphicaly too

Answer 17

awesome. thanks for the help, have a bit better understanding of this now. hopefully i can just get through this calculus class and them fully learn it using resources other than the POS textbook this class uses.

Answer 18

Allright, all the best luck to you. Ask again if you need help!

Answer 19

I have another question I do need help with