Exercise 17 Page 645

Let's consider a cone with radius r and height h. The volume of this cone is one-third the area of the base multiplied by the height.

We want to double the volume of the cone by changing either its radius or its height. Let's find the new dimensions of the cone separately.

Changing the Height

In this part, we will keep the same radius but we will consider a new height h_1. The volume of this new cone is the one shown below. V_1 = 1/3π r^2 h_1 Since we want the volume of the original cone to be doubled, we have that V_1 = 2V. Let's substitute the volume of the original cone and solve the resulting equation for h_1.

V_1 = 2V

SubstituteII

V_1= 1/3π r^2 h_1, V= 1/3π r^2h

1/3π r^2 h_1 = 2( 1/3π r^2h)

▼

Solve for h_1

Multiply

Multiply

1/3π r^2 h_1 = 2/3π r^2h

MultEqn

LHS * 3=RHS* 3

π r^2 h_1 = 2π r^2h

DivEqn

.LHS /π r^2.=.RHS /π r^2.

h_1 = 2h

In consequence, if we want to double the volume of the cone by changing only the height, we must double the height.

Changing the Radius

Now we will keep the original height but we will consider a new radius r_2. With this new radius, the volume of the new cone is given by the following formula. V_2 = 1/3π r_2^2 h Again, since we want to double the volume of the original cone, we set the equation V_2 = 2V and we solve it for r_2.

V_2 = 2V

SubstituteII

V_2= 1/3π r_2^2 h, V= 1/3π r^2h

1/3π r_2^2 h = 2( 1/3π r^2h)

▼

Solve for r_2

Multiply

Multiply

1/3π r_2^2 h = 2/3π r^2h

MultEqn

LHS * 3=RHS* 3

π r_2^2 h = 2π r^2h

DivEqn

.LHS /π h.=.RHS /π h.

r_2^2 = 2r^2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(r_2^2) = sqrt(2r^2)

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

sqrt(r_2^2) = sqrt(2)* sqrt(r^2)

SqrtPowToNumber

sqrt(a^2)=a

r_2 = sqrt(2)r

The final equation implies that to double the volume of the original cone by changing only the radius, the new radius must be sqrt(2) times the original radius.