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Sagot :
Answer:
Ds/dt = 1,25 in/minute
Step-by-step explanation:
s is side of the cube
V(c) = s³
Differentiation on both sides of the equation with respect to time.
DV(c) / dt = 2*s* Ds/dt (1)
In that equation:
s = 2
DV(c) /dt = 5
By subtitution in equation (1)
5 = 2*s*Ds/dt
2*2*Ds/dt = 5
Ds/dt = 5/4 Ds/dt = 1,25 in/minute
The rate of change of the side length of the cube with respect to time, in inches per minute, at the moment when S = 2 inches is [tex]\dfrac{5}{12} \: \rm in^3/min[/tex]
How to calculate the instantaneous rate of growth of a function?
Suppose that a function is defined as;
[tex]y = f(x)[/tex]
Then, suppose that we want to know the instantaneous rate of the growth of the function with respect to the change in x, then its instantaneous rate is given as:
[tex]\dfrac{dy}{dx} = \dfrac{d(f(x))}{dx}[/tex]
Let the rate of change of volume V with respect to time t is given by:
[tex]\dfrac{dV}{dt}[/tex]
And let the rate of change of volume S with respect to time t is given by:
[tex]\dfrac{dS}{dt}[/tex]
The relation between V and S is [tex]V = S^3[/tex]. Using this value, and the chain rule of differentiation, we get:
[tex]\dfrac{dV}{dt} = \dfrac{dS^3}{dt} = 3S^2\dfrac{dS}{dt} = 3S^2\dfrac{dS}{dt}\\\\\dfrac{dS}{dt} = \dfrac{1}{3S^2} \dfrac{dV}{dt}[/tex]
At S = 2, we are given that: [tex]\dfrac{dV}{dt} = 5 \: \rm in^3/min[/tex]
Putting these values in the equation for rate of S, we get:
[tex]\dfrac{dS}{dt} = \dfrac{1}{3S^2} \dfrac{dV}{dt}\\\\\dfrac{dS}{dt} = \dfrac{1}{3(2)^2}\times 5 = \dfrac{5}{12} \: \rm in^3/min[/tex]
Thus, the rate of change of the side length of the cube with respect to time, in inches per minute, at the moment when S = 2 inches is [tex]\dfrac{5}{12} \: \rm in^3/min[/tex]
Learn more about instantaneous rate of change here:
https://brainly.com/question/4746888
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