1) each side of a square is increasing at a rate of 6cm/s. At what rate is the area of the square is 16 cm^2?

2)the radius of a sphere is increasing at a rate of 4mm/s.How fast is the volume increasing when the diameter is 80mm?
3)A particle moves along the curve y= (1+x^3)^1/2.As it reaches the point (2,3), the y-coordinate is increasing at a rate of 4 cm/s.How fast is the x-coordinate of the point changing at that instant?

1) To find the rate at which the area of a square is changing, you need to apply the chain rule from calculus. The area of a square is given by A = s^2, where s is the length of each side.

First, differentiate both sides of the equation with respect to time (t):
dA/dt = d(s^2)/dt

Using the chain rule, differentiate the right side of the equation:
dA/dt = 2s * ds/dt

You are given that ds/dt = 6 cm/s, and you need to find the rate at which the area is changing when the area is 16 cm^2. Substitute these values into the equation:
dA/dt = 2(16) * 6 cm^2/s

Simplifying, you get:
dA/dt = 192 cm^2/s

So, the rate at which the area of the square is increasing is 192 cm^2/s.

2) The volume of a sphere is given by V = (4/3)πr^3, where r is the radius of the sphere.

To find the rate at which the volume is changing, differentiate both sides of the equation with respect to time:
dV/dt = d((4/3)πr^3)/dt

Using the chain rule, differentiate the right side of the equation:
dV/dt = (4/3)(3πr^2) * dr/dt

You are given that dr/dt = 4 mm/s, and you need to find the rate at which the volume is changing when the diameter is 80 mm. The radius (r) is half the diameter, so r = 40 mm. Substitute these values into the equation:
dV/dt = (4/3)(3π(40^2)) * 4 mm^3/s

Simplifying, you get:
dV/dt = 25600π mm^3/s

So, the volume is increasing at a rate of 25600π mm^3/s when the diameter is 80 mm.

3) To find how fast the x-coordinate of a point is changing, you need to differentiate the equation of the curve with respect to time (t) using the chain rule.

Given the equation y = (1 + x^3)^(1/2), you want to find dx/dt when y = 3 and dy/dt = 4 cm/s.

First, differentiate both sides of the equation with respect to t:
dy/dt = d((1 + x^3)^(1/2))/dt

Using the chain rule, differentiate the right side of the equation:
dy/dt = (1/2)(1 + x^3)^(-1/2) * d(1 + x^3)/dt

You are given that dy/dt = 4 cm/s. Also, at the point (2, 3), y = 3. Substitute these values into the equation and solve for dx/dt:
4 cm/s = (1/2)(1 + 2^3)^(-1/2) * d(1 + 2^3)/dt

Simplifying, you get:
4 cm/s = (1/2)(27)^(-1/2) * d(9)/dt

Evaluating the derivatives, you get:
4 cm/s = (1/2)(3/√27) * d(9)/dt

Simplifying further, you get:
4 cm/s = (1/2)(1/√3) * d(9)/dt

Multiplying both sides by √3 and solving for d(9)/dt, you get:
d(9)/dt = (4 cm/s)(2√3) = 8√3 cm/s

So, the x-coordinate of the point is increasing at a rate of 8√3 cm/s at the instant when the y-coordinate is increasing at a rate of 4 cm/s.