A Parametric Surface Question: x = u, y = v, z = u^2 - v^2 - 4

[senolasan]

Hi everyone, I was solving this question however I have some doubts about the correctness of my answer:

Question:
Given a surface having the following parametric equations;
x=u
y=v
z=u²-v²-4
Determine:
1)Point Q, intersection of the surface to the positive x-axis,
2)A Cartesian equation for the plane tangent at Q to the surface

My answer for the first question: if the surface intersects the positive x-axis,
then: y=0 and z=0,

then: v=0
and
u²-v²-4=0 → u²=4 → u=2 since the surface intersects the positive x-axis.

Since u=2 and v=0, we have:
x=2
y=0
z=0
So the point Q=(2,0,0)

My answer for the second question: Since
x=u and y=v, then:
z=x²-y²-4 → "x²-y²-z-4=0" is the Cartesian equation for the plane tangent at Q to the surface.

You would be really helpful if you could point out any mistakes that I did.

 

[Deleted member 4993]

Hi everyone, I was solving this question however I have some doubts about the correctness of my answer:

Question:
Given a surface having the following parametric equations;
x=u
y=v
z=u²-v²-4
Determine:
1)Point Q, intersection of the surface to the positive x-axis,
2)A Cartesian equation for the plane tangent at Q to the surface

My answer for the first question: if the surface intersects the positive x-axis,
then: y=0 and z=0,

then: v=0
and
u²-v²-4=0 → u²=4 → u=2 since the surface intersects the positive x-axis.

Since u=2 and v=0, we have:
x=2
y=0
z=0
So the point Q=(2,0,0)

My answer for the second question: Since
x=u and y=v, then:
z=x²-y²-4 → "x²-y²-z-4=0" → That is not an equation of a plane.

is the Cartesian equation for the plane tangent at Q to the surface.

You would be really helpful if you could point out any mistakes that I did.

A tangent line at (x_o,y_o) in the form of y = mx + b to curve y = f(x) must satisfy:

f'(x_o) = m

What would be equivalent condition/s for tangent plane?

 

[senolasan]

A tangent line at (x_o,y_o) in the form of y = mx + b to curve y = f(x) must satisfy:

f'(x_o) = m

What would be equivalent condition/s for tangent plane?

After some searching, I found that the Cartesian equation for the tangent plane at point Q(2,0,0) to the surface is:
z-z_Q = f_x(x_Q,y_Q)(x-x_Q)+f_y(x_Q,y_Q)(y-y_Q)
z-0 = f_x(2,0)(x-2)+f_y(2,0)(y-0)
Since f_x(2,0) = u² = 2² = 4 and f_y(2,0) = -(v²) = -(0²) = 0
Then z = 4(x-2)+0.y
So the Cartesian equation is 4x-8-z=0 ?