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Anni [7]
3 years ago
9

Traces of surface x2 + y2 − z2 = 1. Determine the equation for the family of traces in x = n.

Mathematics
1 answer:
pickupchik [31]3 years ago
6 0

Answer:

\mathbf{y^2 -z^2 =1- n^2}

Step-by-step explanation:

Give that:

the surface equation is  x^2 +y^2 -z^2 =1

from the family of traces in x = n given that x^2 +y^2 -z^2 =1 , the equation can be represented as :

n^2 +y^2 -z^2 =1

\mathbf{y^2 -z^2 =1- n^2}

This represents a family of hyperbola for all values of n expects that n = ± 1

So, if n = ± 1,

Then

y^2 - z^2 = 0

(y-z) (y+z) = 0

y = ± z

So, for n = ± 1, it is a pair of line for y = z, y = -z

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Example 2:
Montano1993 [528]
Taking this example into account, we can see that setting the first value equal to 1, we obtain that F(x)=0.5x+1=4 and x=6. Using this information, we find that F(x+1)=0.5(x+1)+1=0.5(6+1)+1=4.5. It shows that when x is positive, the succussive terms are increasing.

Referring to that finding, if we set initial value less than zero, which means that we are solving 0.5x+1<0 and taking a number in the interval of the solution, which means x ∈ (- ∞, -20). Setting x=-19, we find that F(x)=0.5x+1=-19 and x=-40. In the next iteration, F(x+1)=0.5(x+1)+1=0.5(1-40)+1=-18.5. In the next iteration, F(x+2)=0.5(x+2)+1=0.5(2-40)+1=-18. By this way, we find that even if the initial value is less than zero, value of the successive iterations is increasing. 

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Setting the initial value equal to 2, we find that g(x)=-x+2=2 and x=0. The next iteration is g(x+1)=-(x+1)+2=1. In this case, the interations are also decreasing. 
If we set the initial value equal to 1, we find that g(x)=-x+2=1 and x=1. In the next iteration, g(x+1)=-(x+1)+2=0 and the iterations are decreasing. 
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