Consider the implicit function y(x) defined by the equation

[tuablink]

Homework Statement

Consider the implicit function y(x) defined by the equation
cos(x-y)=xy for x>0.5
a.find the smallest two positive values x>0.5 for which y(x)=0
b.find the global minimum of y(x) for x>0.5
c.call the numbers you found inpart(b) x1 and x2 find
integral of y(x)dx from x1 to x2

Homework Equations

The Attempt at a Solution

a.One of the two positive values is pi/2 and another one is 3*pi/2
b.I could not figure this out.
c.- 2*cos(y) – pi^2*y

 

[Mark44]

Are what you show in your attempts at a solution the answers in the back of the book? I don't see how you could get something for part c (which depends on part b), if you couldn't figure out part b.

For part b, a maximum or minimum will occur at a point where y'(x) = 0 or is undefined or at an endpoint of the domain. Have you tried differentiating your equation implicitly to find y'(x)?

 

[tuablink]

I did part a. in matlab
Here is the code:
cos(x-y) = x*y
y(x) = 0, cos(x) = 0
solve('cos(x)=0')
ans = pi/2
Here is the code for part c.
int('cos(x-y)-x*y',pi/2,3*pi/2)
ans = - 2*cos(y) – pi^2*y
So, did I do part a. and c. correctly?
I still don't get how to do part b.

 

[Mark44]

What are you going to do in a test situation where you don't have access to matlab? This problem can be done without technology more advanced than pencil and paper.

For part a, if y(x) = 0, substitute 0 for y in the equation cos(x - y) = xy, and find the two smallest x values that are larger than .5.

For part b, this is what I said earlier.

For part b, a maximum or minimum will occur at a point where y'(x) = 0 or is undefined or at an endpoint of the domain. Have you tried differentiating your equation implicitly to find y'(x)?

When you get y'(x), we can work on part c.

 

[tuablink]

I differentiate the equation and I got:
y'(x)=sin(x-y)-x
Substituting y=0 into the equation:
y'(x)=sin(x)-x
I got 0 for the answer but I think it is wrong because it has to be greater than 0.5

 

[Mark44]

I differentiate the equation and I got:
y'(x)=sin(x-y)-x

This is not right. You are forgetting that you need to use the chain rule and the product rule.

Substituting y=0 into the equation:
y'(x)=sin(x)-x
I got 0 for the answer but I think it is wrong because it has to be greater than 0.5

 

[tuablink]

I redo the problem and I got

 

[tuablink]

I redo the problem and I got
y'(x)=-y-sin(x-y)-sin(x-y)+x
y'(x)=-sin(x)-sin(x)+x
y'(x)=-2sin(x)+x

 

[tuablink]

Am I on the right track?

 

[Mark44]

I redo the problem and I got
y'(x)=-y-sin(x-y)-sin(x-y)+x
y'(x)=-sin(x)-sin(x)+x
y'(x)=-2sin(x)+x

Not even close. You should not be starting with y'(x) = <whatever>.

Starting with cos(x - y) = xy, your next equation should have on the left, the derivative with respec to x of cos(x - y). On the right, this equation should have the derivative with respect to x of xy.

To differentiate cos(x - y), you need to use the chain rule. To differentiate xy, you need the product rule.

 

[Mark44]

Also, you have posted the same problem in another thread. Don't do this.

 

[tuablink]

I think I did it right this time.
-sin(x-y)+cos(x-y)=x+y

 

[Mark44]

No, both sides are incorrect.
The derivative with respect to x of cos(x - y) is -sin(x - y) times the derivative with respect to x of x - y. (chain rule)

For the derivative with respect to x of xy, you need the product rule.
d/dx(xy) = x*d/dx(y) + y*d/dx(x). Can you finish this and the one above?

 

[tuablink]

The left hand side is -sin(x-y)*1
The right hand side is y

 

[tuablink]

How can I solve this?

Mod note: merging threads
--Redbelly98

Homework Statement

The derivative of cos(x-y)=xy is -sin(x-y)*(1+dy/dx)=y+x*dy/dx

Homework Equations

The Attempt at a Solution

I simplified a little and I got
-sin(x-y)-dy/dxsin(x-y) = y+x*dy/dx
I got stuck

 

[Mark44]

The left hand side is -sin(x-y)*1
The right hand side is y

No, both are wrong. Did you read what I wrote? Namely, d/dx(xy) = x*d/dx(y) + y*d/dx(x). That's not equal to y.

Go back in your calculus book and review the section on implicit differentiation.

 

[tuablink]

This should be the right derivative:
y'(x)=-y-sin(x-y)/sin(x-y)+x
What do I do next?

 

[Mark44]

It's not right, but it's close.

cos(x - y) = xy
d/dx(cos(x - y)) = d/dx(xy)
-sin(x - y)*(1 - y') = xy' + y
y'sin(x - y) - xy' = y + sin(x -y)
y'(sin(x - y) - x) = y + sin(x - y)
y' = [y + sin(x - y)]/[sin(x - y) - x]

b.find the global minimum of y(x) for x>0.5
c.call the numbers you found inpart(b) x1 and x2 find
integral of y(x)dx from x1 to x2

A minimum or maximum can occur only at a number x at which the derivative is zero or undefined, or at an endpoint of the domain.

Find all numbers x > .5 at which the derivative y' is zero or undefined.

 

[tuablink]

What did I do wrong?

 

[Mark44]

I don't know. I showed you how to do it. Compare what you did with what I did and see where they are different.

 

[tuablink]

Once I finish part b. how can I do part c.

 

[Mark44]

It looks to me like you'll have to solve the differential equation dy/dx = [y + sin(x - y)]/[sin(x - y) - x] to get y(x), and then integrate it as described in your problem.

 

[tuablink]

I tried solving it by using MATLAB but it didn't work out. How can I solve it by hand?

 

[Mark44]

You might start with a substitution with the goal of making the differential equation separable. I would start with u = x - y and see if that helps.

 

[tuablink]

I used the substitution and i got:
u=x-y
du/dx=1-dy/dx
dy/dx=1-du/dx
1-du/dx=u
-du/dx=u-1
-du/u-1=dx
integral(-du/u-1)=integral(dx)
-ln(u-1)=x+c
e^ln(u-1)=e^-x-c
u=e^x-c+1

 

[Mark44]

The suggestion that jambaugh and I gave was to use this substitution in your differential equation y' = [y + sin(x - y)]/[sin(x - y) - x], with the goal of making it separable.

I used the substitution and i got:
u=x-y
du/dx=1-dy/dx
dy/dx=1-du/dx
1-du/dx=u
-du/dx=u-1
-du/u-1=dx
integral(-du/u-1)=integral(dx)
-ln(u-1)=x+c
e^ln(u-1)=e^-x-c
u=e^x-c+1

You did not use the substitution in your differential equation, and besides, you have an error. How can you justify this equation in the fourth line? -- 1-du/dx=u

Several of the integrals in the problems you have posted are very difficult or even impossible to do by analytic means. Are you supposed to be using numerical means to solve the integrals, but haven't mentioned it?