Let's solve the problem step by step: --- ## Given: \( f(x, y) = e^{x^2 + y^2} \) Let \( \mathcal{C} \) be the contour curve of \( f \) at level \( e^2 \). --- ### (a) **Find a Cartesian equation for the tangent line \( \ell \) to \( \mathcal{C} \) at the point \( (1,1) \).** #### **Step 1: Find the equation for the contour curve \(\mathcal{C}\)** Since \( f(x, y) = e^{x^2 + y^2} \), the contour at level \( e^2 \) is: \[ e^{x^2 + y^2} = e^2 \implies x^2 + y^2 = 2 \] #### **Step 2: Find the gradient of \( f \) at \( (1, 1) \)** \[ \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] \[ \frac{\partial f}{\partial x} = 2x e^{x^2 + y^2} \] \[ \frac{\partial f}{\partial y} = 2y e^{x^2 + y^2} \] At \( (1, 1) \): \[ \nabla f(1, 1) = (2 \cdot 1 \cdot e^2, 2 \cdot 1 \cdot e^2) = (2e^2, 2e^2) \] The direction of the tangent line is **perpendicular** to the gradient. #### **Step 3: Find the direction vector for the tangent line** The gradient vector is normal to the curve. The tangent vector is perpendicular to \( (2e^2, 2e^2) \). Any vector \( (a, b) \) perpendicular to \( (2e^2, 2e^2) \) must satisfy: \[ 2e^2 a + 2e^2 b = 0 \implies a + b = 0 \] So a direction vector is \( (1, -1) \). #### **Step 4: Equation of the tangent line** The tangent line at \( (1, 1) \) with direction \( (1, -1) \): \[ (x, y) = (1, 1) + t(1, -1) \] \[ x = 1 + t, \quad y = 1 - t \] Eliminate \( t \): \( t = x - 1 \implies y = 1 - (x - 1) = 2 - x \) **Final equation:** \[ \boxed{y = 2 - x} \] --- ### (b) **Sketch the contour curve \(\mathcal{C}\) and tangent line \(\ell\); show intersections with axes.** #### **Contour Curve (\(x^2 + y^2 = 2\)):** - Circle, center \( (0, 0) \), radius \( \sqrt{2} \). **Intersections:** - \( x \)-axis: \( y = 0 \Rightarrow x^2 = 2 \Rightarrow x = \pm\sqrt{2} \) - \( y \)-axis: \( x = 0 \Rightarrow y^2 = 2 \Rightarrow y = \pm\sqrt{2} \) - Points: \( (\sqrt{2}, 0), (-\sqrt{2}, 0), (0, \sqrt{2}), (0, -\sqrt{2}) \) #### **Tangent Line (\(y = 2 - x\)):** - \( x \)-intercept: \( y = 0 \implies x = 2 \implies (2, 0) \) - \( y \)-intercept: \( x = 0 \implies y = 2 \implies (0, 2) \) --- ### (c) **Equation for the tangent plane \( V \) to the graph of \( f \) in \( \mathbb{R}^3 \) at \( (1, 1, e^2) \).** **General tangent plane to \( z = f(x, y) \) at \( (x_0, y_0, z_0) \):** \[ z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \] Previously found: - \( f_x(1, 1) = 2e^2 \) - \( f_y(1, 1) = 2e^2 \) - \( z_0 = e^2 \) **Plug in:** \[ z = e^2 + 2e^2(x - 1) + 2e^2(y - 1) \] \[ z = e^2 + 2e^2x - 2e^2 + 2e^2y - 2e^2 \] \[ z = 2e^2x + 2e^2y - 3e^2 \] **Final equation:** \[ \boxed{z = 2e^2x + 2e^2y - 3e^2} \] --- ### (d) **At which point on the graph of \( f \) is the tangent plane parallel to \( 2e^{2}x - 2e^{2}y - z = 4 \)?** **Hint:** Planes are parallel if their normal vectors are parallel. #### **Normal vector to given plane:** \( 2e^2x - 2e^2y - z = 4 \implies \) normal vector \( (2e^2, -2e^2, -1) \) #### **Normal vector to tangent plane at \( (a, b, f(a, b)) \):** The tangent plane to \( z = f(x, y) \) at \( (a, b, f(a, b)) \) has normal vector: \[ (f_x(a, b), f_y(a, b), -1) \] Where: \[ f_x(a, b) = 2a e^{a^2 + b^2}, \quad f_y(a, b) = 2b e^{a^2 + b^2} \] Set proportional to \( (2e^2, -2e^2, -1) \): \[ (f_x(a, b), f_y(a, b), -1) = \lambda (2e^2, -2e^2, -1) \] So: \[ 2a e^{a^2 + b^2} = 2\lambda e^2 \implies a e^{a^2 + b^2} = \lambda e^2 \] \[ 2b e^{a^2 + b^2} = -2\lambda e^2 \implies b e^{a^2 + b^2} = -\lambda e^2 \] \[ \Rightarrow a = -b \] Let \( b = -a \). Now, plug into \( a e^{a^2 + b^2} = \lambda e^2 \): But \( a^2 + b^2 = a^2 + (-a)^2 = 2a^2 \): \[ a e^{2a^2} = \lambda e^2 \] \[ b e^{2a^2} = -\lambda e^2 \] So, at \( (a, -a) \), the tangent plane is parallel to the given plane. **For a specific point, choose any \( a \neq 0 \), e.g., \( a = 1 \):** \[ \boxed{(1, -1, e^{1^2 + (-1)^2}) = (1, -1, e^{2})} \] Or in general, **all points** \( (a, -a, e^{2a^2}) \), \( a \neq 0 \). --- ## **Final Answers** --- **(a)** \(\boxed{y = 2 - x}\) **(b)** (Description for sketch: Circle \( x^2 + y^2 = 2 \), tangent line \( y = 2 - x \), intersections: circle at \( (\pm\sqrt{2}, 0), (0, \pm\sqrt{2}) \); tangent line at \( (2, 0), (0, 2) \).) **(c)** \(\boxed{z = 2e^2x + 2e^2y - 3e^2}\) **(d)** \(\boxed{(a, -a, e^{2a^2}),\, a \neq 0}\) — for example, \( (1, -1, e^2) \)