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Magazine Chapter 9: Problem 31
Find the directional derivatives(s) of \(f(x, y)=x+y^{2}\) at \((3,4)\) in the direction of a tangent vector to the graph of \(2 x^{2}+y^{2}=9\) at \((2,1)\).
Short Answer
Expert verified
The directional derivative is \( \frac{-31}{\sqrt{17}} \).
Step by step solution
01
Compute the Gradient of f
First, we need to find the gradient \(abla f\) of the function \(f(x, y) = x + y^2\). The gradient is computed as:\[abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = (1, 2y)\.\]Substitute \((x, y) = (3, 4)\) into \(abla f\):\[abla f(3, 4) = (1, 8)\.\]
02
Calculate the Partial Derivatives of the Implicit Function
The graph of the equation \(2x^2 + y^2 = 9\) is given, and a tangent vector can be found by using implicit differentiation to find the tangent plane.Differentiate both sides with respect to \(x\) :- Differentiate \(2x^2\): \(\frac{d}{dx}(2x^2) = 4x\).- Differentiate \(y^2\) with respect to \(x\): \(2y\cdot\frac{dy}{dx}\).
03
Solve for the Derivative dy/dx
Set up the equation from implicit differentiation:\[4x + 2y\cdot \frac{dy}{dx} = 0\].Solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = -\frac{4x}{2y} = -\frac{2x}{y}\].Substitute \((x, y) = (2, 1)\) to find the slope at this point:\[\frac{dy}{dx} = -\frac{2 \cdot 2}{1} = -4\].
04
Find a Tangent Vector
With the slope \(-4\), we can create a tangent vector to the curve at the point \((2, 1)\). A direction vector can be given by \((1, -4)\), which lies along the tangent line.
05
Normalize the Direction Vector
Normalize the direction vector \((1, -4)\) by dividing by its magnitude to find a unit vector.Compute the magnitude:\[\|\mathbf{v}\| = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}\].The unit vector \(\mathbf{u}\) is:\[\mathbf{u} = \left(\frac{1}{\sqrt{17}}, -\frac{4}{\sqrt{17}}\right)\].
06
Compute the Directional Derivative
The directional derivative \(D_\mathbf{u} f\) in the direction of the unit vector \(\mathbf{u}\) is given by:\[D_\mathbf{u} f = abla f \cdot \mathbf{u} = (1, 8) \cdot \left(\frac{1}{\sqrt{17}}, -\frac{4}{\sqrt{17}}\right)\].Calculate the dot product:\[D_\mathbf{u} f = \frac{1}{\sqrt{17}} - \frac{32}{\sqrt{17}} = \frac{-31}{\sqrt{17}}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
A gradient vector is essential in multivariable calculus as it shows how a function changes at a given point. For any function like our example, \( f(x, y) = x + y^2 \), the gradient, \( abla f \), consists of all its partial derivatives.
- The partial derivative with respect to \( x \) here is \( \frac{\partial f}{\partial x} = 1 \), as differentiating \( x \) results in 1.
- The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 2y \), because the derivative of \( y^2 \) is \( 2y \).
Implicit Differentiation
When dealing with equations defining curves implicitly, such as \( 2x^2 + y^2 = 9 \), direct differentiation is tricky. Implicit differentiation helps find derivatives with respect to \( x \) without explicitly solving for \( y \).
- Differentiating \( 2x^2 \) with respect to \( x \) gives \( 4x \).
- Differentiating \( y^2 \) using the chain rule: it becomes \( 2y \cdot \frac{dy}{dx} \).
Unit Vector
To find directional derivatives, the direction must be a unit vector, meaning it has a magnitude of 1. The tangent vector \( (1, -4) \) has to be normalized.
- Calculate the magnitude: \( \|\mathbf{v}\| = \sqrt{1^2 + (-4)^2} = \sqrt{17} \).
- Divide each component by this magnitude to make the vector a unit vector.
Dot Product
The dot product is a crucial operation to compute directional derivatives. It combines vectors via multiplication and addition, providing a single value. Here, it's used to calculate how the function \( f \) changes in a given direction.
- The directional derivative \( D_\mathbf{u} f \) in the direction of \( \mathbf{u} \) is \( abla f \cdot \mathbf{u} \).
- Calculate by multiplying corresponding components of \( abla f \) and \( \mathbf{u} \), then adding: \[ (1, 8) \cdot \left( \frac{1}{\sqrt{17}}, -\frac{4}{\sqrt{17}} \right) = \frac{1}{\sqrt{17}} - \frac{32}{\sqrt{17}} \].
- This simplifies to \( \frac{-31}{\sqrt{17}} \), representing the rate of change of \( f \) in this direction.
