[tex]\textit{volume of a cube}\\\\ V=s^3 ~~ \begin{cases} s=side \end{cases}\hspace{5em}\cfrac{dV}{dt}=\stackrel{ \textit{chain rule} }{3s^2\cdot \cfrac{ds}{dt}}\implies \cfrac{1}{3s^2}\cdot \cfrac{dV}{dt}=\cfrac{ds}{dt} \\\\\\ \left. \cfrac{ds}{dt} \right|_{s=6}\implies \cfrac{1}{3(6)^2}\cdot \cfrac{dV}{dt}\implies \cfrac{1}{3(6)^2}\cdot 3.5\implies \cfrac{7}{216}~\cfrac{cm}{s}[/tex]
Answer:
[tex]\sf \dfrac{7}{216} \; cm/s\approx 0.0324\;cm/s[/tex]
Step-by-step explanation:
In calculus, a derivative represents the rate of change of a function with respect to its independent variable.
Connected rates of change refer to the relationship between the rates at which two quantities are changing with respect to each other. It involves analysing how changes in one variable affect the rate of change in another variable, often through the use of derivatives.
Some situations have a number of linked variables, like length, surface area and volume or distance, speed and acceleration. By understanding the rate of change of one of these linked variables and the equations that establish connections between them, the chain rule can be applied to determine the rate of change of the other variables.
The formula for the volume of a cube is:
[tex]\boxed{V = s^3}[/tex]
where V is the volume, and s is the side length.
First find the rate of change of V with respect to s by differentiating the function for volume:
[tex]\dfrac{\text{d}V}{\text{d}s}=3s^2\; \sf cm^3/cm[/tex]
[tex]\dfrac{\text{d}V}{\text{d}s}=3s^2\; \sf cm^2[/tex]
If the volume of a cube is increasing at a rate of 3.5 cm³/s, then the rate of change of V with respect to t (time) is:
[tex]\dfrac{\text{d}V}{\text{d}t}=3.5\; \sf cm^3/s[/tex]
To find the rate of change of the side of the base, we need to find the rate of change of s with respect to t, ds/dt.
To do this, we can use the chain rule:
[tex]\dfrac{\text{d}s}{\text{d}t}=\dfrac{\text{d}s}{\text{d}V} \times \dfrac{\text{d}V}{\text{d}t}[/tex]
Substitute the values to find an equation for ds/dt:
[tex]\dfrac{\text{d}s}{\text{d}t}=\dfrac{1}{3s^2\; \sf cm^2} \times 3.5\; \sf cm^3/s[/tex]
[tex]\dfrac{\text{d}s}{\text{d}t}=\dfrac{7}{6s^2}\; \sf cm/s[/tex]
To find the rate of change of the side of the base when its length is 6cm, substitute s = 6 into ds/dt:
[tex]\dfrac{\text{d}s}{\text{d}t}\;\text{at}\;s=6: \quad \dfrac{\text{d}s}{\text{d}t}=\dfrac{7}{6(6)^2}=\dfrac{7}{216}\; \sf cm/s[/tex]
Therefore, the rate of change of the side of the base when its length is 6 cm is:
[tex]\boxed{\sf \dfrac{7}{216} \; cm/s\approx 0.0324\;cm/s}[/tex]
