Problem 14
Question
find the gradient vector of the given function at the given point \(\mathbf{p}\). Then find the equation of the tangent plane at \(\mathbf{p}\) \((\) see Example 1). $$ f(x, y)=\frac{x^{2}}{y}, \mathbf{p}=(2,-1) $$
Step-by-Step Solution
Verified Answer
The gradient vector is \( (-4, -4) \) and the tangent plane equation is \( z = -4x - 4y \).
1Step 1: Find Partial Derivatives
To find the gradient vector, we first need the partial derivatives of the function \( f(x, y) = \frac{x^2}{y} \). The partial derivative with respect to \( x \) is \( f_x = \frac{\partial}{\partial x} \left( \frac{x^2}{y} \right) = \frac{2x}{y} \). The partial derivative with respect to \( y \) is \( f_y = \frac{\partial}{\partial y} \left( \frac{x^2}{y} \right) = -\frac{x^2}{y^2} \).
2Step 2: Evaluate the Gradient at Point \( \mathbf{p} \)
The point given is \( \mathbf{p} = (2, -1) \). Substitute \( x = 2 \) and \( y = -1 \) into the partial derivatives: \( f_x(2, -1) = \frac{2 \times 2}{-1} = -4 \) and \( f_y(2, -1) = -\frac{2^2}{(-1)^2} = -4 \). The gradient vector at \( \mathbf{p} \) is \( abla f(2, -1) = (-4, -4) \).
3Step 3: Gradient Vector Significance
The gradient vector \( abla f(2, -1) = (-4, -4) \) represents the direction of the steepest ascent of the function at the point \( \mathbf{p} = (2, -1) \).
4Step 4: Find Tangent Plane Equation
The general form of the tangent plane at \( \mathbf{p} = (a, b) \) is \( z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \). First, calculate \( f(2, -1) = \frac{2^2}{-1} = -4 \). The tangent plane equation becomes \( z = -4 + (-4)(x - 2) + (-4)(y + 1) = -4 - 4(x - 2) - 4(y + 1) \).
5Step 5: Simplify Tangent Plane Equation
Simplify the equation found in the previous step: \( z = -4 - 4x + 8 - 4y - 4 = -4x - 4y + 0 \). Therefore, the equation of the tangent plane is \( z = -4x - 4y \).
Key Concepts
Partial DerivativesTangent Plane EquationSteepest Ascent Direction
Partial Derivatives
Partial derivatives are derivatives of a multivariable function taken with respect to one variable while keeping the other variables constant. This concept extends the idea of derivative from single-variable calculus to functions of more than one variable. For instance, given the function \[ f(x, y) = \frac{x^{2}}{y} \],we seek to understand how changes in either \(x\) or \(y\) could alter the value of \(f(x, y)\).
- The partial derivative with respect to \(x\), denoted as \(f_x\), examines how \(f\) changes when \(x\) changes slightly, maintaining \(y\) constant: \(f_x = \frac{2x}{y}\).
- The partial derivative with respect to \(y\), denoted as \(f_y\), examines how \(f\) changes when \(y\) changes slightly, maintaining \(x\) constant: \(f_y = -\frac{x^2}{y^2}\).
Tangent Plane Equation
In multivariable calculus, the tangent plane at a point \(\mathbf{p}(a, b)\) on a surface described by a function \(f(x, y)\) is the plane that "just touches" the surface at that point. It is similar to how a tangent line relates to a curve.
For the function \( f(x, y) = \frac{x^2}{y} \), the tangent plane equation at a specific point \( \mathbf{p} = (2, -1) \) can be determined using the formula:\[ z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \],where:
The tangent plane is a powerful tool in linear approximations and can offer valuable insights into the behavior of the surface near \(\mathbf{p}\). It acts as a "flat" predictor of the function's behavior around a localized region.
For the function \( f(x, y) = \frac{x^2}{y} \), the tangent plane equation at a specific point \( \mathbf{p} = (2, -1) \) can be determined using the formula:\[ z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \],where:
- \(f(a, b)\) is the height of the surface at \(\mathbf{p}\).
- \(f_x(a, b)\) and \(f_y(a, b)\) are the gradients of \(f\) with respect to \(x\) and \(y\) at \(\mathbf{p}\).
The tangent plane is a powerful tool in linear approximations and can offer valuable insights into the behavior of the surface near \(\mathbf{p}\). It acts as a "flat" predictor of the function's behavior around a localized region.
Steepest Ascent Direction
The steepest ascent direction is the path along which the function experiences its maximum rate of increase. This concept can be visualized as the direction you would take to climb a hill from any given point, seeking the steepest upward path.
For a multivariable function \(f(x, y)\), this direction is given by the gradient vector \(abla f\). At \(\mathbf{p} = (2, -1)\), the gradient vector is \((-4, -4)\). This tells us that the direction of steepest ascent is towards the vector \((-4, -4)\), a joint movement that increases \(x\) and \(y\) in the negative directions simultaneously.
For a multivariable function \(f(x, y)\), this direction is given by the gradient vector \(abla f\). At \(\mathbf{p} = (2, -1)\), the gradient vector is \((-4, -4)\). This tells us that the direction of steepest ascent is towards the vector \((-4, -4)\), a joint movement that increases \(x\) and \(y\) in the negative directions simultaneously.
- Magnitude: The magnitude or length of \(abla f\) reveals how steep the function is climbing in that direction. A larger magnitude means a steeper ascent.
- Orthogonality: The gradient is perpendicular to the level curve passing through that point, implying no change in function value, hence capturing maximum directional change.
Other exercises in this chapter
Problem 14
Find the global maximum value and global minimum value of \(f\) on \(S\) and indicate where each occurs. \(f(x, y)=x^{2}-6 x+y^{2}-8 y+7 ;\) \(S=\left\\{(x, y):
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If \(z=x y+x+y, x=r+s+t\), and \(y=r s t\), find \(\left.\frac{\partial z}{\partial s}\right|_{r=1, s=-1, t=2}\)
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Find a point on the surface \(z=2 x^{2}+3 y^{2}\) where the tangent plane is parallel to the plane \(8 x-3 y-z=0\).
View solution Problem 14
Find the indicated limit or state that it does not exist. \(\lim _{(x, y) \rightarrow(0,0)} x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}}\)
View solution