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The volume of a right circular cone with radius r and height h is V = (pi r^2 h)/3. a....

Question:

The volume of a right circular cone with radius r and height h is {eq}\displaystyle V=\frac{\pi r^{2}h}{3} {/eq}

a. Approximate the change in the volume of the cone when the radius changes from r = 6.7 to r = 7.2 and the height changes from h = 4.30 to h = 4.25.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.17 to r = 6.14 and the height changes from h = 14.0 to h = 13.95.

Volume:

A quantity that represents the space occupy by the variable or coordinates on the plane by the closed surface is known as volume. Its measurable unit is the cube of the units. For example, the volume of the cube is three times of side of the cube.

Answer and Explanation:

(a)

GIVEN DATA

  • The volume of right circular cone is: {eq}V = \dfrac{1}{3}\pi {r^2}h {/eq}
  • The initial value of radius is: {eq}{r_1} = r = 6.7 {/eq}
  • The final value of radius is: {eq}{r_2} = r = 7.2 {/eq}
  • The initial value of height is: {eq}{h_1} = h = 4.30 {/eq}
  • The final value of height is: {eq}{h_2} = h = 4.25 {/eq}


The expression for change in radius is

{eq}\Delta r = {r_2} - {r_1} {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} \Delta r &= 7.2 - 6.7\\ &= 0.5 \end{align*} {/eq}


The expression for change in height is

{eq}\Delta h = {h_2} - {h_1} {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} \Delta h &= 4.25 - 4.30\\ &= - 0.05 \end{align*} {/eq}


The expression for change in volume is

{eq}dV = \dfrac{{\partial V}}{{\partial r}}\Delta r + \dfrac{{\partial V}}{{\partial h}}\Delta h {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} dV &= \dfrac{{\partial \left( {\dfrac{1}{3}\pi {r^2}h} \right)}}{{\partial r}}\left( {0.5} \right) + \dfrac{{\partial \left( {\dfrac{1}{3}\pi {r^2}h} \right)}}{{\partial h}}\left( { - 0.05} \right)\\ &= \dfrac{2}{3}\pi rh\left( {0.5} \right) - \dfrac{1}{3}\pi {r^2}\left( {0.05} \right) \end{align*} {/eq}


Initial value of radius and height is{eq}r = 6.7 {/eq} and {eq}h = 4.30 {/eq} respectively


Substitute the value and solve the above expression

{eq}\begin{align*} dV &= \dfrac{2}{3}\pi \left( {6.7} \right)\left( {4.30} \right)\left( {0.5} \right) - \dfrac{1}{3}\pi {\left( {6.7} \right)^2}\left( {0.05} \right)\\ &= \dfrac{1}{3}\pi \left( {28.81 - 2.24} \right)\\ &= 27.80 \end{align*} {/eq}


Thus the change in volume of cone is {eq}27.80 {/eq}


(b)

  • The volume of right circular cone is: {eq}V = \dfrac{1}{3}\pi {r^2}h {/eq}
  • The initial value of radius is: {eq}{r_1} = r = 6.17 {/eq}
  • The final value of radius is: {eq}{r_2} = r = 6.14 {/eq}
  • The initial value of height is: {eq}{h_1} = h = 14 {/eq}
  • The final value of height is: {eq}{h_2} = h = 13.95 {/eq}


The expression for change in radius is

{eq}\Delta r = {r_2} - {r_1} {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} \Delta r &= 6.14 - 6.17\\ &= - 0.03 \end{align*} {/eq}


The expression for change in height is

{eq}\Delta h = {h_2} - {h_1} {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} \Delta h &= 13.95 - 14.0\\ &= - 0.05 \end{align*} {/eq}


The expression for change in volume is

{eq}dV = \dfrac{{\partial V}}{{\partial r}}\Delta r + \dfrac{{\partial V}}{{\partial h}}\Delta h {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} dV &= \dfrac{{\partial \left( {\dfrac{1}{3}\pi {r^2}h} \right)}}{{\partial r}}\left( { - 0.03} \right) + \dfrac{{\partial \left( {\dfrac{1}{3}\pi {r^2}h} \right)}}{{\partial h}}\left( { - 0.05} \right)\\ &= - \dfrac{2}{3}\pi rh\left( {0.03} \right) - \dfrac{1}{3}\pi {r^2}\left( {0.05} \right) \end{align*} {/eq}


Initial value of radius and height is {eq}r = 6.17 {/eq} and {eq}h = 14 {/eq} respectively


Substitute the value and solve the above expression

{eq}\begin{align*} dV &= - \dfrac{2}{3}\pi \left( {6.17} \right)\left( {14} \right)\left( {0.03} \right) - \dfrac{1}{3}\pi {\left( {6.17} \right)^2}\left( {0.05} \right)\\ &= - \dfrac{1}{3}\pi \left( {5.1828 +1.9034} \right)\\ &= - 7.42 \end{align*} {/eq}


Thus the change in volume of cone is {eq}7.42 {/eq} and negative sign shows the decrease in volume


Learn more about this topic:

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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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