The volume of a cube is increasing at a rate of 7 cm3/minute. Express the width of the cube, x, in centimeters, as a function of time t in minutes.
Thank you kindly
v = x^3
dv/dx = 3 x^2
dv/dt = 3 x^2 dx/dt
7 = 3 x^2 dx/dt
so
x^2 dx/dt = 7/3
finishing it off,
x^2 dx = 7/3 dt
1/3 x^3 = 7/3 t
x^3 = 7t
x = ∛(7t)
To express the width of the cube, x, as a function of time, t, we need to understand the relationship between the volume of the cube and its width.
The volume of a cube can be expressed as V = x^3, where x is the width of the cube.
Based on the information given, we know that the volume of the cube is increasing at a rate of 7 cm^3/minute. This can be represented as dV/dt = 7 cm^3/minute.
To find the derivative of the volume with respect to time, we can differentiate the volume equation with respect to t:
dV/dt = d/dt(x^3)
Using the power rule of differentiation, we get:
dV/dt = 3x^2 * dx/dt
Now we can substitute the given rate of change, dV/dt = 7, into the equation:
7 = 3x^2 * dx/dt
Since we are trying to express x as a function of t, we can write dx/dt as x'(t):
7 = 3x^2 * x'(t)
Now, we can isolate x in the equation:
7/(3x^2) = x'(t)
To express x as a function of t, we need to integrate both sides with respect to t:
∫ 7/(3x^2) dt = ∫ x'(t) dt
Let's integrate both sides:
(7/3) ∫ x^-2 dx = ∫ dx/dt dt
(7/3) * (-x^-1) = x + C
Now, let's rearrange the equation to solve for x:
-7/(3x) = x + C
To find the value of the constant, C, we need additional information. If we have an initial condition, such as the initial width of the cube, we can substitute it into the equation. Without an initial condition, we cannot determine the exact value of C.
So, the final equation expressing the width of the cube, x, as a function of time, t, is given by:
-7/(3x) = x + C
Remember to substitute a value for C to get a specific solution.