Solving the Discontinuity of y=sqrt(25-x^2)

In summary: So, the piecewise-defined function y = sqrt(25-x^2) -5<x<0 and y = sqrt(25-x^2) 0=<x<5 are both solutions of dy/dx=-x/y on the interval (-5,5).
  • #1
vipertongn
98
0

Homework Statement


y = sqrt(25-x^2) -5<x<0
y = -sqrt(25-x^2) 0=<x<5

dy/dx= -x/y

Why is this not a solution on(-5,5) when...

y = sqrt(25-x^2)
y = -sqrt(25-x^2)
are solutions? kinda confused...

Could someone explain to me step by step how to solve this...I spent an hour trying to understand what they are asking for and solving the problem...haven't gotten anywhere
 
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  • #2


vipertongn said:

Homework Statement


y = sqrt(25-x^2) -5<x<0
y = -sqrt(25-x^2) 0=<x<5

dy/dx= -x/y

Why is this not a solution on(-5,5) when...

y = sqrt(25-x^2)
y = -sqrt(25-x^2)
are solutions? kinda confused...

Could someone explain to me step by step how to solve this...I spent an hour trying to understand what they are asking for and solving the problem...haven't gotten anywhere

I'm confused as well, you didn't state a problem. Are you trying to show that

dy/dx = -x/y

is satisfied by the given equations? If so, it's very straightforward... differentiate (implicitly) the given equations and you can see it has the same form.
 
  • #3


I want to know why it isn't a solution in that piecewise combination
 
  • #4


vipertongn said:

Homework Statement


y = sqrt(25-x^2) -5<x<0
y = -sqrt(25-x^2) 0=<x<5
The function above isn't continuous at x = 0.
vipertongn said:
dy/dx= -x/y

Why is this not a solution on(-5,5) when...

y = sqrt(25-x^2)
y = -sqrt(25-x^2)
are solutions? kinda confused...
You're leaving out some information, it seems. The general solution of the DE dy/dx = -x/y is y = +/-sqrt(C - x^2). An initial condition, which you don't include, would enable you to solve for C, and determine which square root is the unique solution.
vipertongn said:
Could someone explain to me step by step how to solve this...I spent an hour trying to understand what they are asking for and solving the problem...haven't gotten anywhere
Please give us the exact wording of the problem, and we can go from there.
 
  • #5


this is what it exactly says... that
y=phi1(x)=sqrt(25-x^2) and y=phi2(z)=-sqrt(25=x^2) are solutions of dy/dx=-x/y on the interval (-5,5). nd explain why the piece wise solution

y = sqrt(25-x^2) -5<x<0
y = -sqrt(25-x^2) 0=<x<5

isn't a solution of the DE on the interval (-5,5)
 
  • #6


Any solution of the diff. equation has to be differentiable at every point in the open interval (-5, 5). Your piecewise-defined function is neither continuous nor differentiable at x = 0. The two functions phi1(x) and phi2(x) are continuous and differentiable at every point of (-5, 5).
 

Related to Solving the Discontinuity of y=sqrt(25-x^2)

What is the formula for y=sqrt(25-x^2)?

The formula for y=sqrt(25-x^2) is used to represent a semi-circle of radius 5 centered at the origin on a Cartesian plane. It is also known as the upper half of the circle with equation x^2 + y^2 = 25.

Why is there a discontinuity in y=sqrt(25-x^2)?

The discontinuity in y=sqrt(25-x^2) occurs at the endpoints of the semi-circle, where x=5 and x=-5. This is because when x=5 or x=-5, the value of the expression 25-x^2 becomes 0, making it impossible to take the square root of a negative number.

How can the discontinuity in y=sqrt(25-x^2) be solved?

The discontinuity in y=sqrt(25-x^2) can be solved by splitting the function into two separate equations, one for the upper half of the circle (x≤-5 and x≥5) and one for the lower half (-5

What is the domain and range of y=sqrt(25-x^2)?

The domain of y=sqrt(25-x^2) is all real numbers between -5 and 5, inclusive. This is because the square root function is only defined for non-negative numbers, and the expression 25-x^2 must be greater than or equal to 0. The range of y=sqrt(25-x^2) is all real numbers greater than or equal to 0, as the output of the square root function cannot be negative.

Can the discontinuity in y=sqrt(25-x^2) be removed or fixed?

No, the discontinuity in y=sqrt(25-x^2) cannot be completely removed or fixed. However, it can be mitigated by using a piecewise function or by limiting the domain of the function to exclude the endpoints. This is a common approach in mathematics and allows us to work around the discontinuity without changing the overall shape or meaning of the function.

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