A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by V = (x^2)h cm^3. Find the rate at which the volume of the box is changing when the edge length of the base is 10 cm, the edge length of the base is increasing at a rate of 3 cm/min, the height of the box is 5 cm, and the height is decreasing at a rate of 1 cm/min.
just use the product rule
v = x^2 h
dv/dt = 2xh dx/dt + x^2 dh/dt
Now just plug in the given values.
To find the rate at which the volume of the box is changing, we can use the chain rule from calculus.
The volume of the box is given by V = (x^2)h cm^3. Let's find the derivative of V with respect to time.
dV/dt = dV/dx * dx/dt + dV/dh * dh/dt
First, let's find dV/dx, the derivative of V with respect to x. We can differentiate V = (x^2)h with respect to x using the power rule:
dV/dx = 2xh
Next, let's find dx/dt, the rate at which the edge length of the base is increasing. Given that it is increasing at a rate of 3 cm/min, dx/dt = 3 cm/min.
Similarly, let's find dV/dh, the derivative of V with respect to h:
dV/dh = x^2
Next, let's find dh/dt, the rate at which the height of the box is decreasing. Given that it is decreasing at a rate of 1 cm/min, dh/dt = -1 cm/min (since it is decreasing).
Now we can substitute the given values and calculate the rate at which the volume of the box is changing:
dV/dt = (2xh * dx/dt) + (x^2 * dh/dt)
= (2 * 10 * 5 * 3) + (10^2 * -1)
= (100) + (-100)
= 0 cm^3/min
Therefore, the rate at which the volume of the box is changing is 0 cm^3/min.
To find the rate at which the volume of the box is changing, we will need to use the chain rule of differentiation.
Let's start by differentiating the volume equation with respect to time t on both sides:
dV/dt = d/dt(x^2 * h)
Now, using the chain rule, we can differentiate each term separately.
dV/dt = d/dx(x^2 * h) * dx/dt + d/dh(x^2 * h) * dh/dt
Since we are given the values for dx/dt and dh/dt, we can substitute them into the equation.
dx/dt = 3 cm/min (rate at which the edge length of the base is increasing)
dh/dt = -1 cm/min (rate at which the height is decreasing)
Now, let's find the partial derivatives d/dx(x^2 * h) and d/dh(x^2 * h).
d/dx(x^2 * h) = 2xh (using the power rule for differentiation)
d/dh(x^2 * h) = x^2 (since h is treated as a constant when differentiating with respect to x)
Substituting these values into the equation, we have:
dV/dt = (2xh * dx/dt) + (x^2 * dh/dt)
Now let's substitute the given values into the equation:
x = 10 cm (edge length of the base)
h = 5 cm (height of the box)
dx/dt = 3 cm/min (rate at which the edge length of the base is increasing)
dh/dt = -1 cm/min (rate at which the height is decreasing)
dV/dt = (2(10)(5) * 3) + (10^2 * -1)
Simplifying:
dV/dt = (100 * 3) + (100 * -1)
dV/dt = 300 - 100
dV/dt = 200 cm^3/min
Therefore, the volume of the box is changing at a rate of 200 cm^3/min when the given conditions are met.