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?
Just a note of caution.
If this is a regular Calculus course, your questions will not become any easier than these.
1. V = (4/3)π r^3
dV/dt = 4πr^2 dr/dt
for your given data
dV/dt = 4π(64)(14)
= 3584π inches^3/min
SA = 4πr^2
dSA/dt = 8πr dr/dt
again , plug in your given
dSA/dt = 8π(8)(14)
= 896π inches^2/min
y = 4 - x^2
dy/dt = -2x dx/dt
but we need dy/dt = dx/dt, so
dx/dt = -2x dx/dt
divide by dx/dt
1 = -2x
x = -1/2
y = 4 - (-1/2)^2
= 4 - 1/4
= 15/4
the point is (-1/2, 15/4)
1. a) Well, let's start by finding the formula for the volume of a sphere. It's V = (4/3)πr³. Now, we want to find dV/dt (the rate at which the volume is changing) when the radius is 8 in.
We can use the chain rule to solve this. We have dV/dt = dV/dr * dr/dt. The first part, dV/dr, is simply 4πr². The second part, dr/dt, is 14 in/min.
Substituting the values, we get dV/dt = (4π * (8 in)²) * (14 in/min).
b) To find the rate at which the surface area is changing, we can differentiate the formula for the surface area of a sphere. It's A = 4πr².
Using the chain rule again, we have dA/dt = dA/dr * dr/dt. The first part, dA/dr, is 8πr. The second part, dr/dt, is 14 in/min.
Substituting the values, we get dA/dt = (8π * (8 in)) * (14 in/min).
2. Well, to find the point where x and y are changing at the same rate, we need to set their derivatives equal to each other. The derivative of y with respect to x is dy/dx = -2x.
So, we have dy/dx = -2x = 1 (since x and y are changing at the same rate).
Solving for x, we get x = -1/2. Substituting this into the equation for y, we get y = 4 - (-1/2)².
Therefore, at the point (-1/2, 15/4), x and y are changing at the same rate.
Remember, math might not always be funny, but don't worry, I'll try to clown around a bit more in my other answers!
To solve these problems, we will use the chain rule and differentiate the given equations. Let's start with problem 1:
a) To find the rate at which the volume is changing, we need to differentiate the volume formula for a sphere with respect to time. The volume formula for a sphere is V = (4/3)πr^3, where V is the volume and r is the radius.
Differentiating both sides of the equation with respect to time t, we get:
dV/dt = (4/3)π * 3r^2 * dr/dt
We are given that dr/dt = 14 in/min, and we need to find dV/dt when r = 8 in. Substituting these values into the equation, we get:
dV/dt = (4/3)π * 3(8^2) * 14
= (4/3)π * 3 * 64 * 14
= (4/3)π * 192 * 14
= (8/3)π * 2688
= 7168π in^3/min
Therefore, the volume is changing at a rate of 7168π in^3/min when the radius is 8 in.
b) To find the rate at which the surface area is changing, we need to differentiate the surface area formula for a sphere with respect to time. The surface area formula for a sphere is A = 4πr^2, where A is the surface area and r is the radius.
Differentiating both sides of the equation with respect to time t, we get:
dA/dt = 4π * 2r * dr/dt
Substituting the given values, we get:
dA/dt = 4π * 2(8) * 14
= 4π * 16 * 14
= 224π in^2/min
Therefore, the surface area is changing at a rate of 224π in^2/min when the radius is 8 in.
Moving on to problem 2:
To find the point along the curve where x and y are changing at the same rate, we need to find the point where dx/dt (rate of change of x) is equal to dy/dt (rate of change of y).
Given the equation y = 4 - x^2, we need to differentiate both sides of the equation to find dy/dx:
dy/dx = -2x
Now, we set dy/dx equal to dx/dx:
-2x = 1
Solving this equation, we find:
x = -1/2
To find the corresponding y-coordinate, substitute x = -1/2 into the original equation:
y = 4 - (-1/2)^2
= 4 - (1/4)
= 15/4
Therefore, at the point (-1/2, 15/4), x and y are changing at the same rate.
1. a) To determine the rate at which the volume is changing, you need to use the formula for the volume of a sphere, which is V = (4/3)πr^3, where V is the volume and r is the radius.
First, take the derivative of the volume formula with respect to time (t) to find how the volume changes with respect to time:
dV/dt = d/dt[(4/3)πr^3]
Now, we need the relationship between the rate of change of volume and the rate of change of radius. The related rates formula is:
dV/dt = (dV/dr) * (dr/dt)
Here, (dV/dr) represents the derivative of volume with respect to radius (which is 4πr^2) and (dr/dt) represents the rate of change of the radius (given as 14 in/min).
Substitute these values into the formula:
dV/dt = (4πr^2) * (dr/dt) = 4πr^2 * 14
Finally, plug in the radius value of 8 inches into the equation to find the rate of change of volume:
dV/dt = 4π(8^2) * 14
b) To determine the rate at which the surface area is changing, you need to use the formula for the surface area of a sphere, which is A = 4πr^2, where A is the surface area.
Take the derivative of the surface area formula with respect to time (t) to find how the surface area changes with respect to time:
dA/dt = d/dt[4πr^2]
Now, we again use the related rates formula:
dA/dt = (dA/dr) * (dr/dt)
Here, (dA/dr) represents the derivative of surface area with respect to radius (which is 8πr) and (dr/dt) represents the rate of change of the radius (given as 14 in/min).
Substitute these values into the formula:
dA/dt = (8πr) * (dr/dt) = 8πr * 14
Finally, plug in the radius value of 8 inches into the equation to find the rate of change of surface area:
dA/dt = 8π(8) * 14
2. To find the point along the curve where x and y are changing at the same rate, we need to find where the derivative of y with respect to x is equal to 1.
Given y = 4 - x^2, take the derivative of y with respect to x:
dy/dx = d/dx(4 - x^2)
The derivative of 4 is 0, and the derivative of -x^2 is -2x. So, the derivative of the equation is:
dy/dx = -2x
Set dy/dx equal to 1:
-2x = 1
Solve for x:
x = -1/2
Now, substitute the value of x into the original equation to find y:
y = 4 - (-1/2)^2
y = 4 - 1/4
y = 15/4
Therefore, at the point (-1/2, 15/4), x and y are changing at the same rate.