Let's solve each part step by step: --- ### **Given:** A plane has \( x \)-, \( y \)-, and \( z \)-intercepts of **2**, **4**, and **3** respectively. --- ## **Step 1: Find the Equation of the Plane** The general equation for a plane with intercepts is: \[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \] where \( a, b, c \) are the intercepts. Given: \( a=2, b=4, c=3 \) \[ \frac{x}{2} + \frac{y}{4} + \frac{z}{3} = 1 \] We can solve for \( z \) in terms of \( x \) and \( y \) (to write as a function \( f(x, y) \)): \[ \frac{x}{2} + \frac{y}{4} + \frac{z}{3} = 1 \] \[ \frac{z}{3} = 1 - \frac{x}{2} - \frac{y}{4} \] \[ z = 3 \left( 1 - \frac{x}{2} - \frac{y}{4} \right) \] \[ z = 3 - \frac{3x}{2} - \frac{3y}{4} \] So the function is: \[ f(x, y) = 3 - \frac{3}{2}x - \frac{3}{4}y \] --- ## **(a) Find the derivative in the x-direction** This is the partial derivative with respect to \( x \): \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( 3 - \frac{3}{2}x - \frac{3}{4}y \right) \] \[ = 0 - \frac{3}{2} - 0 = -\frac{3}{2} \] --- ## **(b) Find \( D_{(0,1)}f(x, y) \)** This is the **directional derivative** in the direction of the vector \( \mathbf{u} = (0, 1) \). The formula is: \[ D_{\mathbf{u}}f(x, y) = \nabla f(x, y) \cdot \mathbf{u} \] First, compute the gradient: \[ \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = \left( -\frac{3}{2}, -\frac{3}{4} \right) \] So, \[ D_{(0,1)}f(x, y) = \left( -\frac{3}{2}, -\frac{3}{4} \right) \cdot (0, 1) = -\frac{3}{4} \] --- ## **(c) Direction (as a vector) of steepest slope** The direction of steepest slope is **in the direction of the gradient**: \[ \nabla f(x, y) = \left( -\frac{3}{2}, -\frac{3}{4} \right) \] If you want the **unit vector** in this direction: \[ \text{Magnitude} = \sqrt{\left(-\frac{3}{2}\right)^2 + \left(-\frac{3}{4}\right)^2} = \sqrt{\frac{9}{4} + \frac{9}{16}} = \sqrt{\frac{36}{16} + \frac{9}{16}} = \sqrt{\frac{45}{16}} = \frac{3\sqrt{5}}{4} \] So the unit vector is: \[ \left( \frac{-\frac{3}{2}}{\frac{3\sqrt{5}}{4}}, \frac{-\frac{3}{4}}{\frac{3\sqrt{5}}{4}} \right) = \left( -\frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right) \] --- ## **(d) Steepest slope (as a scalar)** The steepest slope is the **magnitude of the gradient** (as calculated above): \[ \boxed{\frac{3\sqrt{5}}{4}} \] --- ### **Summary of Answers** **a)** \( \frac{\partial f}{\partial x} = -\frac{3}{2} \) **b)** \( D_{(0,1)}f(x, y) = -\frac{3}{4} \) **c)** The direction (as a unit vector): \( \boxed{\left( -\frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}} \right)} \) **d)** The steepest slope: \( \boxed{\frac{3\sqrt{5}}{4}} \)

Let's solve each part carefully, with detailed steps and explanations. --- ### **Given:** A plane has x-, y-, and z-intercepts of **2**, **4**, and **3** respectively. --- ### **Step 1: Find the equation of the plane** The intercept form of a plane's equation is: \[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \] where: - \( a = 2 \) (x-intercept) - \( b = 4 \) (y-intercept) - \( c = 3 \) (z-intercept) Substituting: \[ \frac{x}{2} + \frac{y}{4} + \frac{z}{3} = 1 \] Now, **solve for \( z \)** to express \( z \) as a function of \( x \) and \( y \): \[ \frac{z}{3} = 1 - \frac{x}{2} - \frac{y}{4} \] \[ z = 3 \left( 1 - \frac{x}{2} - \frac{y}{4} \right) \] Distribute: \[ z = 3 - \frac{3x}{2} - \frac{3y}{4} \] Define: \[ f(x, y) = z = 3 - \frac{3}{2}x - \frac{3}{4} y \] --- ### **Part (a): Find the partial derivative of \( f \) with respect to \( x \)** \[ f(x, y) = 3 - \frac{3}{2}x - \frac{3}{4} y \] Treat \( y \) as constant: \[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( 3 - \frac{3}{2}x - \frac{3}{4} y \right) \] Derivative: - Derivative of constant \( 3 \) is 0. - Derivative of \( - \frac{3}{2}x \) with respect to \( x \) is \( - \frac{3}{2} \). - Derivative of \( - \frac{3}{4} y \) with respect to \( x \) is 0 (since \( y \) is treated as constant). **Result:** \[ \boxed{ \frac{\partial f}{\partial x} = - \frac{3}{2} } \] --- ### **Part (b): Find the directional derivative of \( f \) at point \( (x, y) \) in the direction \( \mathbf{u} = (0,1) \)** **Step 1:** Compute the gradient \( \nabla f(x, y) \): \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] Already found: \[ \frac{\partial f}{\partial x} = - \frac{3}{2} \] Now, find \( \frac{\partial f}{\partial y} \): \[ f(x, y) = 3 - \frac{3}{2}x - \frac{3}{4} y \] Derivative with respect to \( y \): \[ \frac{\partial f}{\partial y} = - \frac{3}{4} \] **Gradient vector:** \[ \nabla f = \left( - \frac{3}{2}, - \frac{3}{4} \right) \] **Step 2:** Direction vector \( \mathbf{u} = (0, 1) \) **Step 3:** Compute the directional derivative: \[ D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u} = \left( - \frac{3}{2}, - \frac{3}{4} \right) \cdot (0, 1) \] Dot product: \[ = \left( - \frac{3}{2} \times 0 \right) + \left( - \frac{3}{4} \times 1 \right) = 0 - \frac{3}{4} = - \frac{3}{4} \] **Answer:** \[ \boxed{ D_{(0,1)}f = - \frac{3}{4} } \] --- ### **Part (c): Find the **direction** (unit vector) of the **steepest slope** at the point** The steepest slope occurs in the direction of the **gradient vector** \( \nabla f \). - Gradient: \[ \left( - \frac{3}{2}, - \frac{3}{4} \right) \] - **Magnitude** of the gradient: \[ |\nabla f| = \sqrt{\left(- \frac{3}{2}\right)^2 + \left(- \frac{3}{4}\right)^2} \] Compute each term: \[ \left( - \frac{3}{2} \right)^2 = \frac{9}{4} \] \[ \left( - \frac{3}{4} \right)^2 = \frac{9}{16} \] Sum: \[ \frac{9}{4} + \frac{9}{16} = \frac{36}{16} + \frac{9}{16} = \frac{45}{16} \] Square root: \[ |\nabla f| = \sqrt{\frac{45}{16}} = \frac{\sqrt{45}}{4} = \frac{3 \sqrt{5}}{4} \] - **Unit vector in the direction of steepest slope:** \[ \mathbf{u}_{\text{max}} = \frac{\nabla f}{|\nabla f|} = \left( \frac{- \frac{3}{2}}{\frac{3 \sqrt{5}}{4}}, \frac{- \frac{3}{4}}{\frac{3 \sqrt{5}}{4}} \right) \] Simplify each component: For the \( x \)-component: \[ \frac{- \frac{3}{2}}{\frac{3 \sqrt{5}}{4}} = - \frac{3/2}{3 \sqrt{5}/4} = - \frac{3/2 \times 4}{3 \sqrt{5}} = - \frac{(3 \times 4)/2}{3 \sqrt{5}} = - \frac{12/2}{3 \sqrt{5}} = - \frac{6}{3 \sqrt{5}} = - \frac{2}{\sqrt{5}} \] For the \( y \)-component: \[ \frac{- \frac{3}{4}}{\frac{3 \sqrt{5}}{4}} = - \frac{3/4}{3 \sqrt{5}/4} = - \frac{3/4 \times 4}{3 \sqrt{5}} = - \frac{3}{3 \sqrt{5}} = - \frac{1}{\sqrt{5}} \] **Final unit vector:** \[ \boxed{ \left( - \frac{2}{\sqrt{5}}, - \frac{1}{\sqrt{5}} \right) } \] --- ### **Part (d): Find the **steepest slope** (the magnitude of the gradient)** \[ \boxed{ \frac{3 \sqrt{5}}{4} } \] --- ### **Summary of final answers:** | Part | Answer | |--------|---------| | (a) | \( - \frac{3}{2} \) | | (b) | \( - \frac{3}{4} \) | | (c) | \( \boxed{\left( - \frac{2}{\sqrt{5}}, - \frac{1}{\sqrt{5}} \right)} \) | | (d) | \( \boxed{\frac{3 \sqrt{5}}{4}} \) | --- This completes the detailed step-by-step solution!