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प्रश्न
The height of a cone increases by k%, its semi-vertical angle remaining the same. What is the approximate percentage increase (i) in total surface area, and (ii) in the volume, assuming that k is small ?
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उत्तर
Let h be the height, y be the surface area, V be the volume,l be the slant height and r be the radius of the cone.
\[\text { Let } ∆ \text { h be the change in the height }, ∆ \text { r be the change in the radius of base and } ∆ l \text { be the change in the slant height }. \]
\[\text { Semi - vertical angle ramaining the same } . \]
\[ \therefore \frac{∆ h}{h} = \frac{∆ r}{r} = \frac{∆ l}{l}\]
\[\text { Also }, \frac{∆ h}{h} \times 100 = k\]
\[\text { Then }, \frac{∆ h}{h} \times 100 = \frac{∆ r}{r} \times 100 = \frac{∆ l}{l} \times 100 = k . . . \left( 1 \right)\]
\[\left( i \right) \text { Total surface area of the cone, } T = \pi rl + \pi r^2 \]
\[\text { Differentiating both sides w . r . t . r, we get }\]
\[\frac{dT}{dr} = \pi l + \pi r\frac{dl}{dr} + 2\pi r\]
\[ \Rightarrow \frac{dT}{dr} = \pi l + \pi r\frac{l}{r} + 2\pi r \left[ \text { From } \left( 1 \right), \frac{dl}{dr} = \frac{∆ l}{∆ r} = \frac{l}{r} \right] \]
\[ \Rightarrow \frac{dT}{dr} = \pi l + \pi l + 2\pi r \]
\[ \Rightarrow \frac{dT}{dr} = 2\pi\left( l + r \right)\]
\[ \therefore ∆ T = \frac{dT}{dr} ∆ r = 2\pi\left( l + r \right) \times \frac{kr}{100} = \frac{2kr\pi\left( l + r \right)}{100}\]
\[ \therefore \frac{∆ T}{T} \times 100 = \frac{\left( \frac{2kr\pi\left( l + r \right)}{100} \right)}{2\pi r\left( l + r \right)} \times 100 = 2k \] %
\[\text { Hence, the percentage increase in total surface area of cone is } 2k . \] %
\[\left( ii \right) \text { Volume of cone, V } = \frac{1}{3}\pi r^2 h\]
\[\text { Differentiating both sides w . r . t . h, we get }\]
\[\frac{dV}{dh} = \frac{1}{3}\pi r^2 + \frac{1}{3}\pi h2r\frac{dr}{dh}\]
\[ \Rightarrow \frac{dV}{dh} = \frac{1}{3}\pi r^2 + \frac{1}{3}\pi h2r\frac{r}{h} \left[ \text { From} \left( 1 \right), \frac{dr}{dh} = \frac{∆ r}{∆ h} = \frac{r}{h} \right]\]
\[ \Rightarrow \frac{dV}{dh} = \frac{1}{3}\pi r^2 + \frac{2}{3}\pi r^2 \]
\[ \Rightarrow \frac{dV}{dh} = \pi r^2 \]
\[ \therefore ∆ V = \frac{dV}{dh}dh = \pi r^2 \times \frac{kh}{100} = \frac{k\pi r^2 h}{100}\]
\[ \therefore \frac{∆ V}{V} \times 100 = \frac{\left( \frac{k\pi r^2 h}{100} \right)}{\frac{1}{3}\pi r^2 h} \times 100 = 3k \]%
\[\text { Hence, the percentage increase in thevolume of thecone is } 3k .\]%
