# To find the velocity of blood flow using the equation : v ( t ) = 18500 ( 0.25 − r 2 ) on the interval 0 ≤ r ≤ 0.5

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 2.1, Problem 73E

a.

To determine

## To find the velocity of blood flow using the equation : v(t)=18500(0.25−r2)    on the interval 0≤r≤0.5

Expert Solution

The velocity of blood flow when t=0.4 is 4440cm/sec and t=0.4 is 1665cm/sec .

### Explanation of Solution

Given: The particular equation is,

v(t)=18500(0.25r2)    on the interval 0r0.5

Calculation:

Plug t=0.1 in v(t)=18500(0.25r2)

v(0.1)=18500(0.25(0.1)2)v(0.1)=4440cm/sec

Plug t=0.4 in v(t)=18500(0.25r2)

v(0.4)=18500(0.25(0.4)2)v(0.4)=1665cm/sec

Thus, the velocity of blood flow when t=0.4 is 4440cm/sec and t=0.4 is 1665cm/sec .

b.

To determine

### To explain the blood moves through a vein or an artery for the answers found in part a.

Expert Solution

The blood moves through a vein or an artery of radius 0.1cm at velocity of 4440cm/sec while it moves through a vein or an artery of radius 0.4cm at velocity of 1665cm/sec .

### Explanation of Solution

It means that the blood moves through a vein or an artery of radius 0.1cm at velocity of 4440cm/sec while it moves through a vein or an artery of radius 0.4cm at velocity of 1665cm/sec .

c.

To determine

### To determine the values of blood flow velocities for different values of r .

Expert Solution

The values of v(t)=18500(0.25r2) for r=0,0.1,0.2,0.3,0.4,0.5 is shown in table as shown below:

r cmv(t)cm/sec046250.144400.238850.329600.416650.50

### Explanation of Solution

Given:

v(t)=18500(0.25r2)    on the interval 0r0.5

Calculation:

Plug r=0 in v(t)=18500(0.25r2) then,

v(0)=18500(0.2502)=18500(0.25)=4625

Plug r=0.1 in v(t)=18500(0.25r2) then,

v(0.1)=18500(0.250.12)=18500(0.250.01)=18500(0.24)=4440

Plug r=0.2 in v(t)=18500(0.25r2) then,

v(0.2)=18500(0.250.22)=18500(0.250.04)=18500(0.21)=3885

Plug r=0.3 in v(t)=18500(0.25r2) then,

v(0.3)=18500(0.250.32)=18500(0.250.09)=18500(0.16)=2960

Plug r=0.4 in v(t)=18500(0.25r2) then,

v(0.4)=18500(0.250.42)=18500(0.250.16)=18500(0.09)=1665

Plug r=0.5 in v(t)=18500(0.25r2) then,

v(0.5)=18500(0.250.52)=18500(0.250.25)=18500(0)=0

The values of v(t)=18500(0.25r2) for r=0,0.1,0.2,0.3,0.4,0.5 is shown in table as shown below:

r cmv(t)cm/sec046250.144400.238850.329600.416650.50

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