R(t) = –t^2(2t + 4)^2(1 – t)

SOLUTIONS TO HOMEWORK ASSIGNMENT #2, Math 253

1. Find the equation of a sphere if one of its diameters has end points (1, 0, 5) and

(5, −4, 7).

Solution:

The length of the diameter is p

(5 − 1)2 + (−4 − 0)2 + (7 − 5)2 =

√

36 = 6, so the

radius is 3. The centre is at the midpoint ( 1+5

2

,

0−4

2

,

5+7

2

) = (3, −2, 6). Hence, the

sphere is given as (x − 3)2 + (y + 2)2 + (z − 6)2 = 9 .

2. Find vector, parametric, and symmetric equations of the following lines.

(a) the line passing through the points (3, 1,

1

2

) and (4, −3, 3)

Solution:

The vector between two points is ~v = h4 − 3, −3 − 1, 3 −

1

2

i = h1, −4,

5

2

i. Hence

the equation of the line is

Vector form: ~r = ~r0 + t~v = h4, −3, 3i + th1, −4,

5

2

i = h4 + t, −3 − 4t, 3 + 5

2

ti

Parametric form: x = 4 + t, y = −3 − 4t, z = 3 + 5

2

t

Step-by-step explanation:

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.

ok then what happened.