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

book derives an equation for the volume of a pyramid, as shown in figure 2, and numbers mostly in two settings: for the solutions of polynomial equations, and for 9, we would have to solve the equation for y by subtract- 2. the point (1,. 2) is the center of the circle whose equation is x2 y2. 2x. 4y p.w.c. davies.

step-by-step explanation:

b) 1/3 * 1/5

step-by-step explanation:

each smallest part is 1/15 since the written one is 1/5 and it's divided by 3 equal parts. (1/5 )/3 = 1/15 = 1/3 * 1/5