find the tangent plane to the surface z = e^x cos y at the point P(0,0,0)
Question
find the tangent plane to the surface z = e^x cos y at the point P(0,0,0)
Solution
To find the tangent plane to the surface z = e^x cos y at the point P(0,0,1), we first need to find the partial derivatives of the function with respect to x and y.
The partial derivative of z with respect to x is:
dz/dx = e^x cos y
At the point P(0,0,1), this becomes:
dz/dx = e^0 cos 0 = 1
The partial derivative of z with respect to y is:
dz/dy = -e^x sin y
At the point P(0,0,1), this becomes:
dz/dy = -e^0 sin 0 = 0
The equation of the tangent plane at P(0,0,1) is then given by:
z - z0 = dz/dx * (x - x0) + dz/dy * (y - y0)
Substituting the known values, we get:
z - 1 = 1 * (x - 0) + 0 * (y - 0)
Simplifying, we find the equation of the tangent plane to be:
z = x + 1
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