The height of a cone increases by k%, its semi-vertical angle rem… - Vidyadip

Question

The height of a cone increases by k%, its semi-vertical angle remaining the same. What is the approximate percentage increase

In total surface area, and
In the volume, assuming that k is small?

✓

Answer

Let $\triangle\text{h}$ be the change in the height, $\triangle\text{r}$ be the change in the radius of base and $\triangle\text{l}$ be the change in the slant height.

semi-vertical angle remaining the same.

$\therefore\frac{\triangle\text{h}}{\text{h}}=\frac{\triangle\text{r}}{\text{r}}=\frac{\triangle\text{l}}{\text{l}}$

Also, $\frac{\triangle\text{h}}{\text{h}}\times100=\text{k}$

Then, $\frac{\triangle\text{h}}{\text{h}}\times100=\frac{\triangle\text{r}}{\text{r}}\times100=\frac{\triangle\text{l}}{\text{l}}\times100=\text{k}...(1)$

Total surface area of the cone, $\text{T}=\pi\text{rl}+\pi\text{r}^2$

Differentiating both sides w.r.t.

we get

$\frac{\text{dt}}{\text{dr}}=\pi\text{l}+\pi\text{r}\frac{\text{dl}}{\text{dr}}+2\pi\text{r}$

$\frac{\text{dt}}{\text{dr}}=\pi\text{l}+\pi\text{r}\frac{\text{l}}{\text{r}}+2\pi\text{r}\Big[\text{from}(1),\frac{\text{dl}}{\text{dr}}=\frac{\triangle\text{l}}{\triangle\text{r}}=\frac{\text{l}}{\text{r}}\Big]$

$\Rightarrow\frac{\text{dt}}{\text{dr}}=\pi\text{l}+\pi\text{l}+2\pi\text{r}$

$\Rightarrow\frac{\text{dt}}{\text{dr}}=2\pi(\text{l}+\text{r})$

$\therefore\triangle\text{T}=\frac{\text{dt}}{\text{dr}}\triangle\text{r}=2\pi(\text{l}+\text{r})\times\frac{\text{kr}}{100}=\frac{2\text{kr}\pi(\text{l}+\text{r})}{100}$

$\frac{\triangle\text{T}}{\text{T}}\times100=\frac{\Big(\frac{2\text{kr}\pi(\text{l}+\text{r})}{100}\Big)}{2\pi(\text{l}+\text{r})}\times100=2\text{k}\%$

Hence, the percentage increase in total surface area

$\frac{\text{dT}}{\text{dr}}=\pi\text{l}+\pi\text{r}\frac{\text{dl}}{\text{dr}}+2\pi\text{r}$

$\Rightarrow\frac{\text{dT}}{\text{dr}}=\pi\text{l}+\pi\text{r}\frac{\text{l}}{\text{r}}+2\pi\text{r}$

$\Rightarrow\frac{\text{dT}}{\text{dr}}=2\pi(\text{l}+\text{r})$

$\triangle\text{T}=\frac{\text{dT}}{\text{dr}}\triangle\text{r}=2\pi(\text{l}+\text{r})\times\frac{\text{kr}}{100}=\frac{2\text{kr}\pi(\text{l}+\text{r})}{100}$

$\frac{\triangle\text{T}}{\text{T}}\times100=\frac{\frac{2\text{kr}\pi(\text{l}+\text{r})}{100}}{2\pi\text{r}(\text{l}+\text{r})}\times100=2\text{k}\%$

Let v = volume of cone

$=\frac{1}{3}\pi\text{r}^2\text{h}$

$=\frac{1}{3}\pi(\text{h}\tan\alpha)^2\text{h}$

$\text{v}=\frac{\pi}{3}\tan^2\alpha\text{h}^2$

Differentiating it with respect to h treating $\alpha$ as constant,

$\frac{\text{dv}}{\text{dh}}=\pi\tan^2\alpha\times\text{h}$

$\triangle\text{v}=\Big(\frac{\text{dv}}{\text{dh}}\Big)\triangle\text{h}$

$=\pi\tan^2\alpha\text{h}^2\times(0.0\text{kh})$

$\triangle\text{v}=0.0\text{k}\pi\text{h}^3\tan^2\alpha$

Percentage increase in v $=\frac{\triangle\text{v}\times100}{\text{v}}$

$=\frac{0.0\text{k}\pi\text{h}^3\tan^2\alpha\times100}{\frac{\pi}{3}\tan^2\alpha\text{h}^3}$

$=3\text{k}\%$

So, percentage increase in volume $=3\text{k}\%$

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