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?

Explanation

Therefore,
` frac{Delta h}{h}=frac{Delta r}{r}=frac{Delta l}{l} `
Also, ` frac{Delta h}{h} times 100=k `
Then, `frac{Delta h}{h} times 100=frac{Delta r}{r} times 100=frac{Delta l}{l} times 100=k . .(1) `
(i) Total surface area of the cone, `T=pi r l+pi r^{2}`
Differentiating both sides w, r, t, r, we get `frac{d T}{d r}=pi l+pi r frac{d l}{d r}+2 pi r `
` Rightarrow frac{d T}{d r}=pi l+pi r frac{l}{r}+2 pi r[ { From }(1), frac{d l}{d r}=frac{Delta l}{Delta r}=frac{l}{r}]`
` Rightarrow frac{d T}{d r}=pi l+pi l+2 pi r `
` Rightarrow frac{d T}{d r}=2 pi(l+r) `
Therefore, `Delta T=frac{d T}{d r} Delta r=2 pi(l+r) times frac{k r}{100}=frac{2 k r pi(l+r)}{100} `
Therefore, `frac{Delta T}{T} times 100=frac{(frac{2 k r a(l+r)}{100})}{2 pi r(l+r)} times 100=2 k %`
Hence, the percentage increase in total surface area of cone is `2 k . %`
(ii) Volume of cone, `V=frac{1}{3} pi r^{2} h`
Differentiating both sides w. r. t. h, we get
`frac{d V}{d h}=frac{1}{3} pi r^{2}+frac{1}{3} pi h 2 r frac{d r}{d h}`
` Rightarrow frac{d V}{d h}=frac{1}{3} pi r^{2}+frac{1}{3} pi h 2 r frac{r}{h}[ {From}(1), frac{d r}{d h}=frac{Delta r}{Delta h}=frac{r}{h}]`
` Rightarrow frac{d V}{d h}=frac{1}{3} pi r^{2}+frac{2}{3} pi r^{2} `
` Rightarrow frac{d V}{d h}=pi r^{2}`
Therefore,
` Delta V=frac{d V}{d h} d h=pi r^{2} times frac{k h}{100}=frac{k pi r^{2} h}{100} `
Therefore, `frac{Delta V}{V} times 100=frac{(frac{k pi r^{2} h}{100})}{frac{1}{3} pi r^{2} h} times 100=3 k^{2}`
Hence, the parcentage increase in thevolume of thecone is `3 k . %`