What is the equation of the tangent plane to a surface at a specific point?

[Pengwuino]

Ok i need a lil help here.. i got this question I am not sure how to do

Find the equation of the tangent plane to the surface 7z+7=x(e^y)cos(z) at the point (7,0,0). Any know this?

 

[dextercioby]

The equation of the plane is x=0...(the plane yOz)...

Compute the normal vector to the surface in the point (0,0,7).Show it is \vec{i}.

Daniel.

 

[thepatientmental]

The equation of the tangent plane to a surface at a specific point is given by the formula:

z = f(a,b) + fx(a,b)(x-a) + fy(a,b)(y-b)

where f(a,b) is the value of the surface at the point (a,b) and fx(a,b) and fy(a,b) are the partial derivatives of the surface with respect to x and y at the point (a,b), respectively.

In this case, we are given the surface 7z+7=x(e^y)cos(z) and the point (7,0,0). We can find the partial derivatives by taking the derivative of the surface with respect to x and y:

fx = e^y*cos(z)
fy = -x*e^y*sin(z)

At the given point (7,0,0), we have:

f(7,0) = 7(0) + 7 = 7
fx(7,0) = e^0*cos(0) = 1
fy(7,0) = -7*e^0*sin(0) = 0

Substituting these values into the formula for the tangent plane, we get:

z = 7 + (1)(x-7) + (0)(y-0)
or simply,
z = 7 + x - 7
z = x

Therefore, the equation of the tangent plane to the surface 7z+7=x(e^y)cos(z) at the point (7,0,0) is z = x.