1. The radius of a sphere is expanding at a rate of 14 in/min.
a) Determine the rate at which the volume is changing when the radius is 8 in.
b) Determine the rate at which the surface area is changing when the radius is 8 in.
2. A particle moves along a path described by y=4-x^2. At what point along the curve are x and y changing at the same rate?
To find the rate at which the volume and surface area of a sphere are changing, we can use the formulas for volume and surface area of a sphere in terms of its radius.
1a) To determine the rate at which the volume is changing when the radius is 8 inches, we need to use the formula for the volume of a sphere:
V = (4/3)πr^3
Taking the derivative of V with respect to time (t), we can apply the chain rule because r is changing with time:
dV/dt = (4/3)π(3r^2)(dr/dt)
Given that dr/dt = 14 in/min, we substitute r = 8 inches and dr/dt = 14 in/min into the equation to find dV/dt:
dV/dt = (4/3)π(3(8)^2)(14)
= 26752π in^3/min
Therefore, the rate at which the volume is changing when the radius is 8 inches is 26752π in^3/min.
1b) To determine the rate at which the surface area is changing when the radius is 8 inches, we need to use the formula for the surface area of a sphere:
A = 4πr^2
Taking the derivative of A with respect to time (t) and applying the chain rule again, we have:
dA/dt = 4π(2r)(dr/dt)
Given that dr/dt = 14 in/min, we substitute r = 8 inches and dr/dt = 14 in/min into the equation to find dA/dt:
dA/dt = 4π(2(8))(14)
= 896π in^2/min
Therefore, the rate at which the surface area is changing when the radius is 8 inches is 896π in^2/min.
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2. To find the point along the curve where x and y are changing at the same rate, we need to find the points where the derivatives of x and y with respect to t are equal.
The given equation is y = 4 - x^2.
Let's differentiate both sides of this equation with respect to t, assuming that both x and y vary with time:
dy/dt = d(4 - x^2)/dt
= 0 - (2x)(dx/dt)
= -2x(dx/dt)
Since we want x and y to be changing at the same rate, we also need to consider the derivative of x with respect to t, which we'll call dx/dt:
dx/dt = dx/dt
Now we equate the two derivatives to find the point where x and y are changing at the same rate:
-2x(dx/dt) = dx/dt
Rearranging the equation, we get:
-2x(dx/dt) - dx/dt = 0
Factoring out the common term dx/dt, we have:
(-2x - 1)(dx/dt) = 0
To find the points where x and y are changing at the same rate, we solve the equation -2x - 1 = 0 for x:
-2x - 1 = 0
-2x = 1
x = -1/2
Thus, the point along the curve where x and y are changing at the same rate is when x = -1/2. To find the corresponding y-coordinate, substitute x = -1/2 into the equation y = 4 - x^2:
y = 4 - (-1/2)^2
= 4 - 1/4
= 15/4
Therefore, the point along the curve where x and y are changing at the same rate is (-1/2, 15/4).