Homework questions dealing with tangent planes and normal vector

[hwill205]

Hello All,

I need help in my Calc 3 class and I decided to come here for homework help. What I'm looking for is someone to just check my work for a couple of homework problems. I've already done the problems, I would just like my work checked. Anyone who helps, your kindness is greatly appreciated.


Question 1:

Find the equations of the tangent plane and normal line to the given surface at the given point.

z=Log2 (x(y^2)+(x^2)y). So the base of the logarithm function is 2. The point is (1,-2,1)

I rewrote the function as ln(x(y^2) + (x^2)y))/ln(2)

The derivative with respect to x is (y+2x)/((ln2)(xy+x^2))

The derivative with respect to y is (x+2y)/((ln2)(xy+y^2))

The derivative with respect to z is just -1

When you plug in the point (1,-2,1), you get fx=0, fy= -3/((ln2)(2)) and fz=-1

So the equation of the tangent plan is fy(y+2)-(z-1)=0 (it was just easier to use fy instead of the whole expression).

The equation of the normal line is (y+2)/fy=(z-1)/-1, x=1

This is the symmetric equation.


For the second question:

Find the points on the hyperboloid x^2-2y^2-z^2=-2 at which the tangent plane is parallel to the plane 2x-3y+2z+7=0

F(x,y,z)= x^2-2y^2-z^2

The gradient vector for F(x,y,z) is a normal vector for the surface and thus, a normal vector for the tangent plane and for the parallel plane. The gradient vector is:

<2x,-4y,-2z>

so <2x,-4y,-2z>=k<2,-3,3> since <2,-3,3> is a normal vector for the parallel plane.

2x=2k, x=k
-4y=-3k, y=(3/4)k
-2z=2k, z=-k

Plug these values for x,y,z back into the equation for the hyperboloid:

k^2-2((3/4)k)^2-(-k)^2=-2

k^2-(9/8)k^2-k^2=-2

-(9/8)k^2=-2
k=(4/3)

x=4/3
y=1
z=-4/3

The point is (4/3, 1, -4/3)

Can someone tell me here I'm going wrong. I don't think either of these answers are right. Thanks for your help.

 

[hunt_mat]

To find a unit normal to the surface z=\varphi (x,y), you define the following:

<br /> \phi =z-\varphi (x,y)<br />

Then the normal is given by:

<br /> \hat{\mathbf{n}}=\frac{\nabla\phi}{|\nabla\phi |}<br />

It will be easy to compute tangent vectors from this by just looking for 2 linearly independent orthoganal vectors to the normal, the tangent plane at point \mathbf{r}_{0} is given by:

<br /> (\mathbf{r}-\mathbf{r}_{0})\cdot\hat{\mathbf{n}}=0<br />

 

[HallsofIvy]

More simply, two tangent vectors to the surface z= \phi(x,y) are
\vec{i}+ \phi_x\vec{k} and \vec{j}+ \phi_y\vec{k}.

 

[hwill205]

So where is my answer wrong for the first question? I'm still confused...