Acube is expanding and its side is changing at the rate of 2cm/s.When the side is 8cm, find the rate at which:

Question

a) volume is increasing. V=x^3
b) surface area is increasing. A=4x^2

Answer 1

Part a)
V = x²
dV/dx = 3x²
dV/dt = (dV/dx) (dx/dt)
dV/dt = (3 x ²) (2)
dV/dt = 6x²
dV/dt = 6 x 64
dV/dt = 384 cm ³ / s

Part b)
A = 6 x² (there are 6 faces)
A `(x) = 12 x
dA/dt = (12 x) (dx/dt)
dA/dt = (96) (2)
dA/dt = 192 cm ² / s

Answer 2

Given dx/dt = 2 cm/s and x = 8 cm

Differentiate Volume and Surface Area with respect to time

V = x^3
dV/dt = 3x^2 dx/dt
dV/dt = 3(8^2)(2) = 384 cu cm /s, the rate at w/c Volume is increasing

A = 4x^2
dA/dt = 8x dx/dt
dA/dt = 8(8)(2) = 128 sq cm / s, the rate at w/c Surface Area is increasing

Answer 3

a. V = x^3 ....... dx/dt = 2
dV/dt = 3x^2 dx/dt = (3*8^2)2
dV/dt = 384

b. A = 4x^2 ....... dx/dt = 2
dA/dt = 8x dx/dt = 8(8)2
dA/dt = 128

Answer 4

a