The respect to \( x \) equals$$ frac{Delta y}{Delta x}

The average rate of change of one variable (e.g., \( y ext{)} \) with respect to another (e.g., \( x ext{)} \) equals the ratio of the two variables’;changes.$$ ext{average rate of change}= frac{ ext{change in } y}{ ext{change in } x} = frac{Delta y}{Delta x} $$So, if the value of \( y \) changes from \( y_1 \) to \( y_2 \) while the value of \( x \) changes from \( x_1 \) to \( x_2 \) (i.e., over the interval \( x_1, x_2 ext{),} \) then the average rate of change of \( y \) with respect to \( x \);equals$$ frac{Delta y}{Delta x} = frac{y_2 – y_1}{x_2 – x_1} $$If \( y = f(x) ext{,} \) then the average rate of change of \( y \) over the interval \( c, x \);equals$$ frac{y_2 – y_1}{x_2 – x_1} = frac{f(x_2) – f(x_1)}{x_2 – x_1} = frac{f(x) – f(c)}{x – c} $$Note that the volume \( V \) of a cube is a function of the length \( s \) of its;edges.$$ V = f(s) = s^3 $$Therefore, the average rate of change of \( V \) with respect to \( s \) over the interval \( c, s \);equals$$ frac{Delta V}{Delta s} = frac{V_2 – V_1}{s_2 – s_1} = frac{f(s_2) – f(s_1)}{s_2 – s_1} = frac{f(s) – f(c)}{s – c} $$First, given \( c = ???s1??? \) and \( s = ???s2??? ext{,} \) calculate \( f(c) \) and;\( f(s) ext{.} \)$$ f(c) = f(???s1???) = left( ???s1???
ight)^3 = ???s1cubed??? $$$$ f(s) = f(???s2???) = left( ???s2???
ight)^3 = ???s2cubed??? $$Now, use substitution to solve for the average rate of;change.$$ egin{align} ext{average rate of change} &= frac{f(s) – f(c)}{s – c} cr&= frac{left( ???s2cubed???
ight) – left( ???s1cubed???
ight)}{left( ???s2???
ight) – left( ???s1???
ight)} cr&= frac{???diffCubes???}{0.1} cr&= ???ARoC???end{align} $$Therefore, the average rate of change of the cube’s volume with respect to an edge as \( s \) decreases from \( ???s1??? \) to \( ???s2??? \) mm is \( ???ARoC??? ext{ mm}^3 ext{/mm.} \)To solve for the instantaneous rate of change of the volume with respect to an edge, take the limit of the average rate of change formula as \( x \) approaches \( c ext{.} \)$$ egin{align} ext{instantaneous rate of change} &= lim_{x o c} frac{f(x) – f(c)}{x – c} cr&= lim_{x o ???s1???} frac{x^3 – ???s1???^3}{x – ???s1???}end{align} $$\( c \) and \( f(c) \) are known values. Although you do not know the exact values of \( x \) (only that it is very close to \( c ext{)} \) or \( f(x) = x^3 ext{,} \) you can still simplify the fraction by factoring.Use the formula for the difference of cubes, \( a^3 – b^3 = left( a – b
ight) left( a^2 + ab + b^2
ight) ext{,} \) to remove the discontinuity in the limit’s;denominator.$$ egin{align} ext{instantaneous rate of change} &= lim_{x o ???s1???} frac{x^3 – left( ???s1???
ight)^3}{x – ???s1???} cr&= lim_{x o ???s1???} frac{left( x – ???s1???
ight) left( x^2 + ???s1??? x + left( ???s1???
ight)^2
ight)}{x – ???s1???} cr&= lim_{x o ???s1???} frac{left( x – ???s1???
ight) left( x^2 + ???s1??? x + ???s1squared???
ight)}{x – ???s1???} cr&= lim_{x o ???s1???} left( x^2 + ???s1??? x + ???s1squared???
ight) crend{align} $$Finally, take the limit by substituting \( ???s1??? \) for;\( x ext{.} \)$$ egin{align} ext{instantaneous rate of change} &= left( ???s1???
ight)^2 + ???s1??? left( ???s1???
ight) + ???s1squared??? cr&= ???IRoC???end{align} $$Therefore, the instantaneous rate of change of the cube’s volume with respect to an edge when \( s \) equals \( ???s1??? \) mm is \( ???IRoC??? ext{ mm}^3 ext{/mm.} \)