Find an equation of the tangent plane to the surface $z=x^2+y^2$ at the point $(1,2,5)$.

• A. $z=2(x-1)+4(y-2)+5$
• B. $z=2x+4y-5$
• C. $z=2x+4y-1$
• D. $z=2x+4y-5$

Answer: (C)

Verified step by step guidance

1. Step 1: Recall the general formula for the equation of the tangent plane to a surface z = f(x, y) at a point (x₀, y₀, z₀). The formula is given by: z - z₀ = fₓ(x₀, y₀)(x - x₀) + fᵧ(x₀, y₀)(y - y₀), where fₓ and fᵧ are the partial derivatives of f with respect to x and y, respectively.
2. Step 2: Compute the partial derivative of the given surface z = x² + y² with respect to x. This gives fₓ(x, y) = ∂/∂x(x² + y²) = 2x.
3. Step 3: Compute the partial derivative of the given surface z = x² + y² with respect to y. This gives fᵧ(x, y) = ∂/∂y(x² + y²) = 2y.
4. Step 4: Evaluate the partial derivatives at the given point (1, 2). For fₓ(1, 2), substitute x = 1 and y = 2 into fₓ(x, y) = 2x, resulting in fₓ(1, 2) = 2(1) = 2. Similarly, for fᵧ(1, 2), substitute x = 1 and y = 2 into fᵧ(x, y) = 2y, resulting in fᵧ(1, 2) = 2(2) = 4.
5. Step 5: Substitute the values x₀ = 1, y₀ = 2, z₀ = 5, fₓ(1, 2) = 2, and fᵧ(1, 2) = 4 into the tangent plane formula z - z₀ = fₓ(x₀, y₀)(x - x₀) + fᵧ(x₀, y₀)(y - y₀). Simplify the equation to find the tangent plane: z = 2(x - 1) + 4(y - 2) + 5.