Let s be the surface defined by x 2 + 2y 3 + 3z 4 = 6. let t be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−2, uv+1). these two surfaces intersect in a curve c which passes through the point (x, y, z) = (1, 1, 1). find the tangent line to c through (1, 1, 1).

The tangent to through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces and at that point.

Let . Then is the level curve . Recall that the gradient vector is perpendicular to level curves; we have

so that the gradient of at (1, 1, 1) is

For the surface , we have

so that . We can obtain a vector normal to by taking the cross product of the partial derivatives of , and evaluating that product for :

Now take the cross product of the two normal vectors to and :

The direction of vector (24, 8, -8) is the direction of the tangent line to at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by . Then adding (1, 1, 1) shifts this line to the point of tangency on . So the tangent line has equation

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

sadsdda

step-by-step explanation: