An asteroid moves toward the sun along a straight line with a velocity given by v = -(c1+c2/x)½ , where x is the distance from the center of the sun. Use the chain rule for derivatives to show that the asteroid's acceleration is a = -c2 /2x^2.
To find the acceleration of the asteroid, we need to differentiate the velocity function with respect to time. The velocity function is given by v = -(c1 + c2/x)^(1/2).
Let's first find the derivative of v with respect to x:
dv/dx = d/dx[-(c1 + c2/x)^(1/2)]
To use the chain rule, let's define a new function u = c1 + c2/x. Therefore, we have:
u = c1 + c2/x
If we differentiate u with respect to x, we get:
du/dx = d/dx(c1 + c2/x)
Using the power rule for differentiation, the derivative of 1/x with respect to x is -1/x^2. Therefore, we have:
du/dx = d/dx(c1) + d/dx(c2/x) = 0 + c2(-1/x^2) = -(c2/x^2)
Now, let's differentiate v with respect to u. We have:
dv/du = d/dx[-(u)^(1/2)] = -(1/2)(u)^(-1/2)
Substituting u back in, we get:
dv/du = -(1/2)(c1 + c2/x)^(-1/2)
To find dv/dx, we can multiply dv/du by du/dx:
dv/dx = -(1/2)(c1 + c2/x)^(-1/2) * (-(c2/x^2))
Simplifying the expression:
dv/dx = (1/2)(c2/x^2)(c1 + c2/x)^(-1/2)
The acceleration, a, is the derivative of v with respect to time, t, which can be obtained by multiplying dv/dx by dx/dt:
a = dv/dx * dx/dt
To express dx/dt in terms of x, we use the velocity function v = dx/dt:
v = -(c1 + c2/x)^(1/2)
Solving for dx/dt:
dx/dt = -(c1 + c2/x)^(1/2)
Now, let's substitute dv/dx and dx/dt back into the equation for a:
a = (1/2)(c2/x^2)(c1 + c2/x)^(-1/2) * (-(c1 + c2/x)^(1/2))
Simplifying the expression further:
a = -(1/2)(c2/x^2)(c1 + c2/x)^(-1/2)(c1 + c2/x)^(1/2)
Since (c1 + c2/x)^(1/2) and (c1 + c2/x)^(-1/2) cancel out:
a = -(1/2)(c2/x^2)
Which is the given expression for the asteroid's acceleration, a = -c2/2x^2.
Therefore, using the chain rule for derivatives, we have shown that the asteroid's acceleration is a = -c2/2x^2.