Solving the Domain: sqrt(x^2+y^2-1)>0 and 1<= x^2+y^2 <=20

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In summary, the conversation is discussing how to combine the domain into an expression of the form a>b and b<c. The domain is given as sqrt(x^2+y^2-1)>0 and 1<=x^2+y^2<=20, and the question is whether it is possible to convert it into a single expression. The speaker suggests that the first condition implies 1<x^2+y^2, and to satisfy both conditions, the intersection of the implication of the first condition and the second condition must be found. This results in the requirement of 1<x^2+y^2<=20. The conversation concludes with the speaker expressing that this method will make their calculations easier.
  • #1
Kris1
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Im trying to combine this domain into something of the form a>b and b<c however I cannot seem to convert this equation :/

Is it possible to convert this domain into a single expression??

The domain is

sqrt(x^2+y^2-1)>0 and 1<= x^2+y^2 <= 20

any ideas?? I've encountered this problem many times and I am unsure as to how to solve this. Is there a general formula to apply?
 
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  • #2
The first condition:

\(\displaystyle \sqrt{x^2+y^2-1}>0\)

implies:

\(\displaystyle 1<x^2+y^2\)

Because both conditions must be true (as implied by the "and" between them), we find the intersection of the implication of the first condition with the second condition, to find we require:

\(\displaystyle 1<x^2+y^2\le 20\)

in order for both conditions to be satisfied.
 
  • #3
Thanks this makes a lot of sense. Combining two like this makes a lot of sense. This will make all of my further calculations easier thank you :)
 

FAQ: Solving the Domain: sqrt(x^2+y^2-1)>0 and 1<= x^2+y^2 <=20

1. What is the domain for the given equation?

The domain for the given equation is all values of x and y that satisfy the inequality sqrt(x^2+y^2-1)>0 and 1<= x^2+y^2 <=20.

2. How do I graph the solution for this equation?

To graph the solution, you can first graph the two equations sqrt(x^2+y^2-1)=0 and x^2+y^2=20 separately. The first equation represents a circle with a radius of 1 centered at the origin, while the second equation represents a circle with a radius of √20 centered at the origin. The solution will be the shaded region between the two circles, with the origin excluded.

3. Can I use substitution to solve this equation?

No, substitution is not an effective method for solving this equation. Since there are two variables (x and y), it is difficult to isolate one variable and solve for it. Graphing or using calculus methods would be more effective.

4. Is there a specific technique for solving this type of equation?

This is a system of inequalities, and there are various methods for solving these types of equations. One approach could be to find the intersection points of the two circles and then test points within the shaded region to see if they satisfy the given inequality. Another approach could be to use calculus methods to find the maximum and minimum values of the equation and determine the corresponding values of x and y.

5. How can I check if my solution is correct?

You can check your solution by substituting the values of x and y into the original inequality and seeing if it holds true. Additionally, you can graph the solution and see if it matches the shaded region described by the inequality.

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