Question

Determine whether the points p and q lie on the given surface. r(u, v) = u + v, u2 − v, u + v2 p(3, 3, 3), q(4, −2, 10)

Answer 1

Answer:

If  p  and q  lie on the surface , then we should be able to find values of  u  and  v  such that they can satisfy the equation of the surface.

i.e

x = u+v;  

y = $u^2 - v$ and

z = $u + v^2$

For p we have to find u and v such that:

u + v =3 ;                      ......[1]

$u^2 - v$ = 3 ;     .....[2]

$u + v^2$ = 3       .....[3]

Equate [1] and [3] we have;

u + v = $u + v^2$

Subtract u from both sides we get;

u + v -u= $u + v^2-u$

Simplify:

$v = v^2$

or

$v^2 - v = v(v-1) = 0$

⇒ v = 0 and v = 1

To find the value of u substitute the values of v in [1]

for v =0

u + 0 = 3

u = 3

for v = 1

u + 1 = 3

Subtract 3 from both sides we get;

u =  2

Now; substitute the values of u and v in equation [2] to satisfy:

if u = 3 ad v = 0

then;

$u^2 - v$ = 3

$3^2 - 0$ = 3

9 = 3 which is not true

if u =2 and v =1

then;

$2^2 - 1$ = 3

4-1 = 3

3 = 3 which is true.

Therefore, the point p lies on the given surface with u = 2 and v =1

Similarly, for q we have to find u and v such that:

u + v =4 ;                      ......[1]

$u^2 - v$ = -2 ;     .....[2]

$u + v^2$ = 10       .....[3]

Adding first two equation; we get

$u + v +u^2 -v = 4-2$

Simplify:

$u^2+u -2 =0$

or

$(u+2)(u-1)$ = 0

⇒ u = -2 and u = 1

from [1] we have  v = 4- u

For u = -2

then;

v = 4-(-2) = 4+2

v = 6

For u = 1

then;

v = 4-1

v =3

Now; substitute the values of u and v in equation [3] to satisfy:

if u = -2 ad v = 6

then;

$u +v^2$ = 10

$-2 +6^2$ = 10

-2 +36 = 10

34 = 10 which is not true

if u = 1 ad v = 3

then;

$u +v^2$ = 10

$1 +3^2$ = 10

1 +9 = 10

10 = 10 which is not true.

Therefore, the point q lies on the given surface with u=1 and v =3