Consider the hyperbola y = 1/x and think of it as a slide. A particle slides along the hyperbola so that its x-coordinate is increasing at a rate of f(x) units/sec. If its y-coordinate is decreasing at a constant rate of 1 unit/sec, what is f(x)?
-I arrived at the answer f(x)= x^2 but i am not quite sure about my answer, pls help for my assigment.
y = 1/x
dy/dt = -1/x^2 dx/dt
-1 = -1/x^2 dx/dt
dx/dt = x^2
Looks good to me.
To find the rate at which the x-coordinate is changing (f(x)), we can differentiate the equation of the hyperbola with respect to time (t).
Given: y = 1/x
Differentiating both sides with respect to t gives:
dy/dt = d(1/x)/dt
To simplify, let's find dy/dx first:
dy/dx = d(1/x)/dx = -1/x^2
Now, we need to relate dy/dx to dy/dt using the chain rule:
dy/dt = dy/dx * dx/dt
Substituting the values we have:
-1/x^2 = dy/dt * dx/dt
We are given that dy/dt is constant and equal to -1. So, we have:
-1/x^2 = -1 * dx/dt
Simplifying further:
dx/dt = x^2
Therefore, the rate at which the x-coordinate is changing (f(x)) is given by dx/dt = x^2.
Hence, your answer of f(x) = x^2 is correct.