0
$\begingroup$

enter image description here

This is the work I have so far:

$f_x = 2x_0,\ \ \ x_0 = x_0 \ \ \ \ \ \ f_x = 2x_0$

$f_y = 6y_0, \ \ \ \ \ y_0 = 0 \ \ \ \ \ \ f_y = 0$

$f_z = 20z_0, \ \ \ \ \ z_0 = 0 \ \ \ \ \ \ f_z = 0 \\$

$f_x(x-x_0) + f_y(y-y_0) + f_z(z-z_0) = 0$

$2x_0(1 - x_0) = 0$

$2x_0 - 2x_0^2$

$x = 1$

Apparently, the answer is actually e) but I am unsure of what I am doing wrong.

$\endgroup$
  • $\begingroup$ evaluate the derivatives at $ (1, -1, 1 )$ not $( x_0, 0, 0)$ $\endgroup$ – WW1 May 9 '15 at 20:10
  • $\begingroup$ Why would that be so? In my textbook the equation of a tangent plane is written as $f_x(P_0)(x - x_0) + f_y(P_0)(y - y_0) + f_z(P_0)(z - z_0) = 0$. I understand that switching it to (1, -1, 1) will give the right answer, I just don't understand why if the formula given in my textbook calls for $(x_0, 0, 0)$ $\endgroup$ – Robert May 9 '15 at 20:12
  • $\begingroup$ In your textbook's formula, $x_0$ is the $x$-coordinate of the point on the surface given by $f(x,y,z)=0$, and so on for $y_0$ and $z_0$. You are trying to make those variables to be the intersection of the tangent plane with the $x$-axis, quite a different thing. You need to first find the equation of the tangent plane, using the book's formula with $x_0=1,y_0=-1,z_0=1$, and then solve that equation for $x$ after substituting $y=0,z=0$. $\endgroup$ – Rory Daulton May 9 '15 at 20:24
1
$\begingroup$

It is unfortunate that the question named the x-intercept $x_0$ which appears as a symbol in your formula making reference to the point on the ellipsoid. The question might be clearer if they said the x-intercept was at $(a,0,0)$ and asked you to find $a$

you should use $P_0=(x_0, y_0, z_0)=(1,-1,1)$

so the equation of the tangent plane is $2(x-1) -6(y+1)+20(z-1)=0$ and the x-intercept is at $(14,0,0)$

$\endgroup$
  • $\begingroup$ Quick question: shouldn't the 3(z-1) actually be 20(z-1)? $\endgroup$ – Robert May 9 '15 at 21:25
  • $\begingroup$ I got the equation $2(x−1)−6(y+1)+20(z−1)=0. \\ f_x = 2x = 2(1) = 2. \\ f_z = 20z = 20(1) = 2$ $\endgroup$ – Robert May 9 '15 at 21:32
  • $\begingroup$ you are correct about the $20(z-1)$ I have edited my answer accordingly $\endgroup$ – WW1 May 9 '15 at 21:35
  • $\begingroup$ For Future Viewers: I made an error in the original post. The answer was actually e) and the correct equation is 2(x−1)−6(y+1)+20(z−1)=0. $\endgroup$ – Robert May 9 '15 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.