If so, compare dz/dx and dz/dy at that point with
6x+8y-z=25
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kuntilanakAuthor Commented:
it means squared
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kuntilanakAuthor Commented:
so is the derivative of z equals to the tangent of the plane?
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> Which of the following vectors is normal to the surface at the given point?
To find the surface normal:
Method 1. Notice that the surface is just an inverted cone with its vertex at the origin? So the normal is quite intuitively going to be (x,y,-1/2).
Method 2. If you can't quite spot the above, then you can work it out by first parametrising the surface:
p(z,t) = (sqrt(z)*cos(t), sqrt(z)*sin(t), z),
then the surface normal is defined as:
N = dp/dz x dp/dt (where x is the vector cross product).
The result comes to (x,y,-1/2) again.
> 6x+8y-z=25
Are you sure they want you to derive it? It looks more like they're telling you what the tangent plane is, and then expect you to use it. If not though, then the tangent plane at (x_0, y_0) for a surface z=f(x,y) is defined as:
z = f(x_0, y_0) + df/dx(x_0,y_0)(x-x_0) + df/dy(x_0,y_0)(y-y_0)
where f=x^2+y^2, and x_0=3, y_0=4 in your case. So plugging that in and rearranging leads to the required equation..
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kuntilanakAuthor Commented:
>>Are you sure they want you to derive it? It looks more like they're telling you what the tangent plane is, >>and then expect you to use it. If not though, then the tangent plane at (x_0, y_0) for a surface >>z=f(x,y) is defined as:
I don't even know if they want me to derive it or not.... that's what I am asking though
>> what equation is this for
z = f(x_0, y_0) + df/dx(x_0,y_0)(x-x_0) + df/dy(x_0,y_0)(y-y_0)
"the tangent plane at (x_0, y_0) for a surface z=f(x,y)"
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If so, compare dz/dx and dz/dy at that point with
6x+8y-z=25