Chapter 1.5, Problem 122E

(a)

To determine

To find: Thebasic definition of derivative of under root x.

Answer to Problem 122E

Thesphere, cylinder and cone will have equal volume

  ${\text{V}}_{\text{cyl}}{\text{=V}}_{\text{con}}{\text{=V}}_{\text{s}}$ or $\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}={\text{πr}}^{\text{2}}{\text{h}}_1=\frac{\text{1}}{\text{3}}{\text{πr}}^{\text{2}}{\text{h}}_2$ .

Explanation of Solution

Given:

All sphere, cylinder and cone have same radius and same volume.

Concept used:

The volume of the sphere: ${\text{V}}_{\text{s}}\text{=}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}$ .

The volume of the cylinder: ${\text{V}}_{\text{cyl}}{\text{=πr}}^{\text{2}}{\text{h}}_1$ .

The volume of the cone: ${\text{V}}_{\text{con}}\text{=}\frac{\text{1}}{\text{3}}{\text{πr}}^{\text{2}}{\text{h}}_2$ .

Calculation:

The volume of the sphere: ${\text{V}}_{\text{s}}\text{=}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}$ .

The volume of the cylinder: ${\text{V}}_{\text{cyl}}{\text{=πr}}^{\text{2}}{\text{h}}_1$ .

The volume of the cone: ${\text{V}}_{\text{con}}\text{=}\frac{\text{1}}{\text{3}}{\text{πr}}^{\text{2}}{\text{h}}_2$ .

The sphere, cylinder and cone will have equal volume when its moulded to each other the shape of the structure changes volume remain same due to having geometrical shapes different each of the solid geometry have different formula but equal volume.

Hence, the sphere, cylinder and cone will have equal volume

  ${\text{V}}_{\text{cyl}}{\text{=V}}_{\text{con}}{\text{=V}}_{\text{s}}$ or $\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}={\text{πr}}^{\text{2}}{\text{h}}_1=\frac{\text{1}}{\text{3}}{\text{πr}}^{\text{2}}{\text{h}}_2$ .

(b)

To determine

To find:The equation of the cylinder’s height and cone height.

Answer to Problem 122E

The height of the cylinder ${\text{h}}_{\text{1}}\text{=}\frac{\text{4}}{\text{3}}\text{r}$ and height of the cone ${\text{h}}_{\text{2}}\text{=4r}$ .

Explanation of Solution

Given: $\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}{\text{=πr}}^{\text{2}}{\text{h}}_{\text{1}}$ and $\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\text{=}\frac{\text{1}}{\text{3}}{\text{πr}}^{\text{2}}{\text{h}}_{\text{2}}$ .

Concept used:

The volume of the sphere: ${\text{V}}_{\text{s}}\text{=}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}$ .

The volume of the cylinder: ${\text{V}}_{\text{cyl}}{\text{=πr}}^{\text{2}}{\text{h}}_1$ .

The volume of the cone: ${\text{V}}_{\text{con}}\text{=}\frac{\text{1}}{\text{3}}{\text{πr}}^{\text{2}}{\text{h}}_2$ .

Calculation:

  ${\text{V}}_{\text{s}}{\text{=V}}_{\text{cyl}}\Rightarrow \frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}{\text{=πr}}^{\text{2}}{\text{h}}_{\text{1}}$ .

Dividing by ${\text{πr}}^{\text{2}}$ :

  ${\text{h}}_{\text{1}}\text{=}\frac{\text{4}}{\text{3}}\text{r}$ .

  ${\text{ V}}_{\text{s}}{\text{=V}}_{\text{con}}\Rightarrow \frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\text{=}\frac{\text{1}}{\text{3}}{\text{πr}}^{\text{2}}{\text{h}}_{\text{2}}$ .

Dividing by $\frac{1}{3}{\text{πr}}^{\text{2}}$ :

  ${\text{h}}_{\text{2}}\text{=4r}$ .

Hence the height of the cylinder ${\text{h}}_{\text{1}}\text{=}\frac{\text{4}}{\text{3}}\text{r}$ and height of the cone ${\text{h}}_{\text{2}}\text{=4r}$ .