a particle moves along the curve y= sqrt 1+x cubed. As it reaches the point (2,3) the y-corrdinate is increasing at a rate of 4cm/s. How fast is the x-coordinate of the point changing at that instant?
y= sqrt(1+x^3)
dy/dt= 1/2 * 1/sqrt(1+x^3)* 3x^2 dx/dt
dy/dt= 3x^2/2y * dx/dt
dx/dt= 2y dy/dt * 1/3x^2
you are given x, y, dy/dt, solve fod dx/dt.
check my work, I did it in a hurry
hey thanks we are checking it now:)
To find how fast the x-coordinate of the point is changing at a given instant, we need to use the chain rule from calculus. The chain rule states that if y is a function of x, and x is a function of t (time), then dy/dt = dy/dx * dx/dt.
Let's start by finding dy/dx, which is the derivative of y with respect to x. We have the equation y = √(1 + x^3), so we can find the derivative using the power rule and the chain rule.
dy/dx = d/dx (√(1 + x^3))
= (1/2) * (1 + x^3)^(-1/2) * 3x^2
= 3x^2 / (2√(1 + x^3))
Now we have dy/dx, but we need to find dx/dt, which represents the rate at which the x-coordinate is changing with respect to time. Unfortunately, we don't have any information about dx/dt given in the problem.
However, we can use the fact that the particle is moving along the curve to find the relationship between dx/dt and dy/dt. We know that dy/dt = 4 cm/s (the rate at which the y-coordinate is increasing). But dy/dt can also be expressed as dy/dx times dx/dt: dy/dt = dy/dx * dx/dt.
Plugging in the values, we have 4 cm/s = (3x^2 / (2√(1 + x^3))) * dx/dt.
Now we can solve for dx/dt:
dx/dt = (4 cm/s) * (2√(1 + x^3)) / (3x^2)
To find the value of dx/dt when the point reaches (2,3), we substitute x = 2 into the equation:
dx/dt = (4 cm/s) * (2√(1 + 2^3)) / (3 * 2^2)
= (4 cm/s) * (2√9) / 12
= (4 cm/s) * (6/12)
= 2 cm/s
Therefore, the x-coordinate of the point is changing at a rate of 2 cm/s at the instant when the y-coordinate is increasing at a rate of 4 cm/s.