I need help setting up this problem:
A particle moves along the curve
y=sqr(1+x^3). As it reaches the point (2,3) the y-cordinate is increasing at a rate of 4cm/s. How fast is the x-coordinate of the point changing at that instant.
I could only come up with this:
dy/dx=4
Both y and x are functions of t.
dy/dt = (dy/dx)*(dx/dt)
You are given that dy/dt = 4 and, according to my calculations,
dy/dx = [(1/2)/sqr(1+x^3)]*3x^2
= [(1/2)/3]*12 = 2
That means dx/dt = 4/2 = 2
To find how fast the x-coordinate is changing, we need to use the chain rule.
Let's start by differentiating both sides of the equation y = sqrt(1 + x^3) with respect to x.
dy/dx = d/dx(sqrt(1 + x^3))
To simplify this expression, we can rewrite it as dy/dx = (1 + x^3)^(1/2).
Now we have the expression for dy/dx, but we're given that dy/dt = 4 cm/s.
To relate dy/dt with dx/dt, we can use the chain rule again.
dy/dt = dy/dx * dx/dt
Knowing that dy/dt = 4 cm/s, we can substitute it into the equation:
4 cm/s = (1 + x^3)^(1/2) * dx/dt
Now we need to solve for dx/dt.
To do that, we rearrange the equation:
dx/dt = 4 cm/s / (1 + x^3)^(1/2)
So, the rate of change of the x-coordinate at that instant is given by dx/dt = 4 cm/s / (1 + x^3)^(1/2). To find the specific value, substitute the x-coordinate of the point (2, 3) into the equation.