\displaystyle \textbf{Question 1: }~\text{Find the rate of change of the total surface area of a cylinder of radius } \\ r\text{ and height }h,\text{ when the radius varies.}
\displaystyle \text{Answer:}

\displaystyle \text{Let } T \text{ be the total surface area of a cylinder. Then, } T = 2\pi r(r + h)

\displaystyle \text{Since the radius varies, we differentiate the total surface area with respect to } r.

\displaystyle \text{Now, } \frac{dT}{dr} = \frac{d}{dr}\big[2\pi r(r + h)\big]

\displaystyle \Rightarrow \frac{dT}{dr} = \frac{d}{dr}(2\pi r^2) + \frac{d}{dr}(2\pi rh)

\displaystyle \Rightarrow \frac{dT}{dr} = 4\pi r + 2\pi h

\displaystyle \Rightarrow \frac{dT}{dr} = 2\pi(2r + h)

\displaystyle \textbf{Question 2: }~\text{Find the rate of change of the volume of a sphere with respect to its diameter.}
\displaystyle \text{Answer:}

\displaystyle \text{Let } V \text{ and } r \text{ be the volume and diameter of the sphere, respectively. Then, }

\displaystyle V = \frac{4}{3}\pi(\text{radius})^3

\displaystyle \Rightarrow V = \frac{4}{3}\pi\left(\frac{r}{2}\right)^3 = \frac{1}{6}\pi r^3

\displaystyle \Rightarrow \frac{dV}{dr} = \frac{1}{2}\pi r^2

\displaystyle \textbf{Question 3: }~\text{Find the rate of change of the volume of a sphere with respect to its} \\ \text{surface area when the radius is }2\text{ cm.}
\displaystyle \text{Answer:}

\displaystyle \text{Let } V \text{ be the volume of the sphere. Then, } V = \frac{4}{3}\pi r^3

\displaystyle \Rightarrow \frac{dV}{dr} = 4\pi r^2

\displaystyle \text{Let } S \text{ be the total surface area of the sphere. Then, } S = 4\pi r^2

\displaystyle \Rightarrow \frac{dS}{dr} = 8\pi r

\displaystyle \therefore \frac{dV}{dS} = \frac{\dfrac{dV}{dr}}{\dfrac{dS}{dr}}

\displaystyle \Rightarrow \frac{dV}{dS} = \frac{4\pi r^2}{8\pi r} = \frac{r}{2}

\displaystyle \Rightarrow \left(\frac{dV}{dS}\right)_{r=2} = \frac{2}{2} = 1\ \text{cm}

\displaystyle \textbf{Question 4: }~\text{Find the rate of change of the area of a circular disc with respect} \\ \text{to its circumference when the radius is }3\text{ cm.}
\displaystyle \text{Answer:}

\displaystyle \text{Let } A \text{ be the area of the circular disc. Then, } A = \pi r^2

\displaystyle \Rightarrow \frac{dA}{dr} = 2\pi r

\displaystyle \text{Let } C \text{ be the circumference of the circular disc. Then, } C = 2\pi r

\displaystyle \Rightarrow \frac{dC}{dr} = 2\pi

\displaystyle \therefore \frac{dA}{dC} = \frac{\dfrac{dA}{dr}}{\dfrac{dC}{dr}}

\displaystyle \Rightarrow \frac{dA}{dC} = \frac{2\pi r}{2\pi} = r

\displaystyle \Rightarrow \left(\frac{dA}{dC}\right)_{r=3} = 3\ \text{cm}

\displaystyle \textbf{Question 5: }~\text{Find the rate of change of the volume of a cone with respect to the} \\ \text{radius of its base.}
\displaystyle \text{Answer:}

\displaystyle \text{Let } V \text{ be the volume of the cone. Then, } V = \frac{1}{3}\pi r^2 h

\displaystyle \Rightarrow \frac{dV}{dr} = \frac{2}{3}\pi r h

\displaystyle \textbf{Question 6: }~\text{Find the rate of change of the area of a circle with respect to its radius} \\ r\text{ when }r=5\text{ cm.}
\displaystyle \text{Answer:}

\displaystyle \text{Let } A \text{ be the area of the circle. Then, } A = \pi r^2

\displaystyle \Rightarrow \frac{dA}{dr} = 2\pi r

\displaystyle \text{Hence, the rate of change of the area of the circle is } 2\pi r

\displaystyle \text{When } r = 5\ \text{cm},

\displaystyle \left(\frac{dA}{dr}\right)_{r=5} = 2\pi(5) = 10\pi\ \text{cm}^2/\text{cm}

\displaystyle \textbf{Question 7: }~\text{Find the rate of change of the volume of a ball with respect to its radius }r. \\ \text{ How fast is the volume changing with respect to the radius when the radius is }2\text{ cm?}
\displaystyle \text{Answer:}

\displaystyle \text{Let } V \text{ be the volume of the spherical ball. Then, } V = \frac{4}{3}\pi r^3

\displaystyle \Rightarrow \frac{dV}{dr} = 4\pi r^2

\displaystyle \text{Thus, the rate of change of the volume of the sphere is } 4\pi r^2

\displaystyle \text{When } r = 2\ \text{cm},

\displaystyle \left(\frac{dV}{dr}\right)_{r=2} = 4\pi(2)^2 = 16\pi\ \text{cm}^3/\text{cm}

\displaystyle \textbf{Question 8: }~\text{The total cost }C(x)\text{ associated with the production of }x\text{ units of an item is given by} \\ C(x)=0.007x^3-0.003x^2+15x+4000.\text{ Find the marginal cost when }17\text{ units are produced.}
\displaystyle \text{Answer:}

\displaystyle \text{Since the marginal cost is the rate of change of total cost with respect to its output,}

\displaystyle \text{Marginal Cost (MC)} = \frac{dC}{dx}

\displaystyle \Rightarrow \frac{dC}{dx} = \frac{d}{dx}\left(0.007x^3 - 0.003x^2 + 15x + 4000\right)

\displaystyle = 0.021x^2 - 0.006x + 15

\displaystyle \text{When } x = 17,

\displaystyle \text{Marginal Cost (MC)} = 0.021(17)^2 - 0.006(17) + 15

\displaystyle = 6.069 - 0.102 + 15 = \text{Rs } 20.967

\displaystyle \textbf{Question 9: }~\text{The total revenue received from the sale of }x\text{ units of a product is given by } \\ R(x)=13x^2+26x+15.\text{ Find the marginal revenue when }x=7.
\displaystyle \text{Answer:}

\displaystyle \text{Since the marginal revenue is the rate of change of total revenue with respect to its output,}

\displaystyle \text{Marginal Revenue (MR)} = \frac{dR}{dx}

\displaystyle \Rightarrow \frac{dR}{dx} = \frac{d}{dx}(13x^2 + 26x + 15)

\displaystyle = 26x + 26

\displaystyle \text{When } x = 7,

\displaystyle \text{Marginal Revenue (MR)} = 26(7) + 26

\displaystyle = 182 + 26 = \text{Rs } 208

\displaystyle \textbf{Question 10: }~\text{The money to be spent for the welfare of the employees of a firm is} \\ \text{proportional to the rate of change of its total revenue (marginal revenue). If the total } \\ \text{revenue (in rupees) received from the sale of } x\text{ units of a product is given by }R(x)=3x^2+36x+5,\text{ find the marginal revenue when } \\ x=5,\text{ and write which value does the question indicate.}\;[\text{CBSE 2013}]
\displaystyle \text{Answer:}

\displaystyle \text{Since marginal revenue is the rate of change of total revenue with respect} \\ \text{to the number of units sold, we have}

\displaystyle \text{Marginal Revenue (MR)} = \frac{dR}{dx} = 6x + 36

\displaystyle \text{When } x = 5,\ \text{MR} = 6(5) + 36 = 66

\displaystyle \text{Hence, the required marginal revenue is } \text{Rs }66.

\displaystyle \text{It indicates the extra revenue obtained when the number of units sold increases from } 5 \text{ to } 6.

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