Question

The length t, widt w, and height h of a box change with time. At a certain instant the dimensions are f=1 m and w=h=4 m, and t and w are increasing at a rate of 6 m/s while h is decreasing at a rate of 7 m/5. At that instant find the rates at which the following quantities are changing.

(a) The volume=___ m³/s

Answer 1

The rate at which the volume V of a box is changing given the changing lengths t, w, and h is computed using the derivative of the volume formula. At the given instant, the volume is changing at a rate of 114.4 m³/s.

Step-by-step explanation:

The student is inquiring about the rate at which the volume of a box is changing over time given the rates of change of its length, width, and height. To find the rate of change of the volume, we will use the derivative of the volume formula with respect to time.

The volume V of a box is given by the product of its length t, width w, and height h, that is, V = t × w × h. To find the rate at which the volume is changing, we'll differentiate both sides of this equation with respect to time t:

dV/dt = w × h × (dt/dt) + t × h × (dw/dt) + t × w × (dh/dt)

Plugging in the given values and rates, we have:

dV/dt = 4 × 4 × 6 + 1 × 4 × 6 + 1 × 4 × (-7/5).

Therefore, dV/dt = 96 + 24 - 5.6 = 114.4 m³/s. This is the rate at which the volume of the box is changing at the given instant.