A metal cube dissolves in acid such that an edge of the cube decreases by 0.49 mm/min. How fast is the volume?

Question

of the cube changing when the edge is 6.7 mm?
*related rates problem*
please explain steps
any help is greatly appreciated.

Answer 1

Write down all relevant variables, and units:
V = volume in mm^3
t = time in min
x = edge length in mm

What down what you are told:
dx/dt = 0.49
V = x^3

Write down what you want:
dV/dt

To get dV/dt, we need dV/dx and dx/dt. The second one is 0.49, and the first one can be gotten by differentiating the volume equation: dV/dx = 3x^2. Substitute in x = 6.7, and the final answer is
3*6.7^2 * 0.49 = 65.9883 mm^3 / min (decreasing)

Answer 2

permit the length of one fringe of the cube be x. If the quantity of the cube is V, then: a million). V = x^3. Taking the time differential provides: 2). dV/dt = 3x^2(dx/dt). Plugging on your numbers provides: 3). dV/dt = 3(7.5^2)((-0.fifty 8) = -ninety seven.875 cubic mm according to minute.

Answer 3

v=r^3
dr/dt=-.49
we have to find dv/dt
dv/dt=3*r^2*(dr/dt)
=65.9883 cumm/min
thanks & good luck