Consider the surface, s, described by the equation z = x sin(x + y). (a) does (2, −2, 0) belong to this surface? if so, what is the equation of the tangent plane, t, to the surface at this point? (b) suppose r(t) = ha(t), b(t), c(t)i is a vector valued function whose associated space curve, c, is on the surface s. that is, we suppose c(t) = a(t) sin(a(t) + b(t)) for all t ∈ r. show that if r(0) = h2, −2, 0i, then r 0 (0) is parallel to t. (c) we now suppose r 0 (0) 6= 0. is the tangent line to c at r(0) on the plane t? can you infer from this a general fact about space curves on surfaces, tangent lines, and tangent planes?

The answer to this question can be given as:

Tangent line at r(0)[more generally r(to)]  is on tangent plane of S at that point.

Step-by-step explanation:

In option (A):

put x=2 ,y=-2 in Z=x sin(x+y).

Z=2 sin(2+(-2))

Z=2 sin(2-2)

Z=2 sin(0)  

we know that sin o= 0;

so (2,-2,0) ∈ s

dz/dx=sin(x+y)+xcos(x+y)

dz/dx=x cos(x+y)

tangent plane T at (a,-2,0) is:

z-0= dz/dx(2,-2)(x-2)+dz/dy(2,-2)(y+2)

z=(0+2)(x-2)+(2)(y+2)

z=2(x-2)+2(y+2)

z=2x-4+2y+4

z=2x+2y is the plane at(2,-2,0)

In option (B):

r(t) is on s

r'(t) =(a'(t),b'(t),c'(t))

this is  tangent rector on S for Vt

so  r'(0) is  tangent to 5

r'(0) is  parallel to T[tangent plane or 5]

In option (C):

yes.

fact , for any surface S. Choose any curve acts in S

if r'(0)!= 0 or[more generally r'(b)!=0]

=Tangent line at r(0)[more generally r(to)]

is on tangent plane of S at that point.

step-by-step explanation: