Mathematics: 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).

Let s be the surface defined by x 2 + 2y 3 + 3z 4

Ответ:

beckytank6338

12.01.2022

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

Let f(x,y,z)=x^2+2y^3+3z^4 . Then is the level curve f(x,y,z)=6 . Recall that the gradient vector is perpendicular to level curves; we have

$\nabla f(x,y,z)=(2x,6y,12z^2)$

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

$\nabla f(1,1,1)=(2,6,12)$

For the surface , we have

$\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0$

so that $\vec r(1,0)=(1,1,1)$ . We can obtain a vector normal to by taking the cross product of the partial derivatives of $\vec r(u,v)$ , and evaluating that product for u=1,v=0 :

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)$

$\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)$

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

$(2,6,12)\times(1,-1,2)=(24,8,-8)$

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 $t\in\Bbb R$ . Then adding (1, 1, 1) shifts this line to the point of tangency on . So the tangent line has equation