Chain rule derivative applied to an ice cube

[sporus]

Homework Statement

A cubical block of ice is melting in such a way that each edge decreases steadily by 9.8 cm every hour. At what rate is its volume decreasing when each edge is 10 meters long?

Homework Equations

V(t) = (l(t))^3 m^3
l'(t) = 0.098 m/h

The Attempt at a Solution

V'(t) = f'(g)*g(t) chain rule formula, f(g) = g^3, g(t) = l = 10
V'(t) = 3g^2 * 0.098
V'(t) = 3(10)^2 * 0.098
V'(t) = 300 * 0.098
V'(t) = 29.4 m^3/h

 

[Mark44]

Homework Statement

A cubical block of ice is melting in such a way that each edge decreases steadily by 9.8 cm every hour. At what rate is its volume decreasing when each edge is 10 meters long?

Homework Equations

V(t) = (l(t))^3 m^3
l'(t) = 0.098 m/h

The Attempt at a Solution

V'(t) = f'(g)*g(t) chain rule formula, f(g) = g^3, g(t) = l = 10
V'(t) = 3g^2 * 0.098
V'(t) = 3(10)^2 * 0.098
V'(t) = 300 * 0.098
V'(t) = 29.4 m^3/h

My only real quibble is that V'(t) should be negative, since the block of ice is melting.

 

[sporus]

that was the answer, thanks. i don't like doing online assignments because there is no feedback. the website kept telling me i was wrong and i had no idea where

 

[Mark44]

This is how I would do this problem.

V(t) = (x(t))³
V'(t) = 3(x(t))² * x'(t)

x'(t) = -0.098 m/hr is constant.
At some time t = t₀, x(t₀) = 10,
so V'(t₀) = 3 * 10² * -0.098 = -29.4 m³/hr.