# Answer to Question #15838 in Mechanics | Relativity for asif rafiq

Question #15838

let a satellite is moving in an orbit given by the equation x2+y2=4.At what rate the position of the satellite y is changing with respect to x.When the satellite is at the point (2^1/2, 2^1/2)

Expert's answer

Let's find the dependence of y from x:

x²+y²=4 ==> y(x) = √(4-x²).

Now we can find derivative of y(x) with respect to x:

y'(x) = (1/2)*(4-x²)^(-1/2)*(-2x) = -x(4-x²)^(-1/2).

So, when x is equal to √2 (and so does y)

y'(√2) = -√2(4-(√2)²)^(-1/2) = -√2*2^(-1/2) = -√2*(1/√2) = -1.

Therefore, y is changing with the rate of -1 with respect to x when the satellite is at the point (√2,√2).

x²+y²=4 ==> y(x) = √(4-x²).

Now we can find derivative of y(x) with respect to x:

y'(x) = (1/2)*(4-x²)^(-1/2)*(-2x) = -x(4-x²)^(-1/2).

So, when x is equal to √2 (and so does y)

y'(√2) = -√2(4-(√2)²)^(-1/2) = -√2*2^(-1/2) = -√2*(1/√2) = -1.

Therefore, y is changing with the rate of -1 with respect to x when the satellite is at the point (√2,√2).

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