Find the equation of the tangent plane to the surface z e^{4x/17}ln(2y) at the point (-2, 2,0.8659).

To find the equation of the tangent plane to the surface, we need to first find the gradient vector of the surface at the given point. The gradient vector is perpendicular to the tangent plane, so once we have it, we can use it to write the equation of the tangent plane. The surface is given as {eq}f(x, y) = e^{4x/17} ln(2y) {/eq}, so we need to find the partial derivatives of {eq}f(x, y) {/eq} with respect to {eq}x {/eq} and {eq}y {/eq}: {eq}frac{partial f}{partial x} = \frac{4}{17} e^{4x/17} ln(2y) \quad \text{ and} \quad \frac{partial f}{partial y} = \frac{1}{y} e^{4x/17} {/eq} Now we can evaluate these partial derivatives at the given point {eq}(-2, 2, 0.8659) {/eq}: {eq}frac{partial f}{partial x}(-2, 2) = \frac{4}{17} e^{-8/17} ln(4) \approx 0.2400 \quad \text{ and} \quad \frac{partial f}{partial y}(-2, 2) = \frac{1}{2} e^{-8/17} \approx 0.0908 {/eq} So the gradient vector at the point {eq}(-2, 2, 0.8659) {/eq} is: {eq}nabla f(-2, 2) = left\frac{4}{17} e^{-8/17} ln(4), \frac{1}{2} e^{-8/17}, 0right) {/eq} This gradient vector is perpendicular to the tangent plane, so we can use it to write the equation of the tangent plane with the point {eq}(-2, 2, 0.8659) {/eq}. The equation of a plane can be written as: {eq}a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 {/eq} where {eq}(x_0, y_0, z_0) {/eq} is a point on the plane and {eq}(a, b, c) {/eq} is a normal vector to the plane. We can use the gradient vector as the normal vector, and the given point {eq}(-2, 2, 0.8659) {/eq} as a point on the plane. Plugging these values into the equation, we get: {eq}frac{4}{17} e^{-8/17} ln(4) (x + 2) + \frac{1}{2} e^{-8/17} (y - 2) = 0.2400(2.3496) + 0.0908(-0.8659) {/eq} Simplifying, we get the final equation of the tangent plane: {eq}0.693x + 0.045(y - 2) + z - 0.224 = 0 {/eq}