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59. Let P(x, y, z) = real number. Let a be a real number such that the rate of change of P at the point (-2,-2,4) in the direction of (a,−1, a) is 52 dP Let ß be a value such that =-6 at t = 0. dt Find the value of the poduct aß. −−²+tan¯¹ (xyz) +3e*+2, and let r(t) be a space curve such that r(0) = (−2, 7, 4) and r '(0) = (1, ß, –4), where ß is some Yex²

59. Let P(x, y, z) = real number. Let a be a real number such that the rate of change of P at the point (-2,-2,4) in the direction of (a,−1, a) is 52 dP Let ß be a value such that =-6 at t = 0. dt Find the value of the poduct aß. −−²+tan¯¹ (xyz) +3e*+2, and let r(t) be a space curve such that r(0) = (−2, 7, 4) and r '(0) = (1, ß, –4), where ß is some Yex²

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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### Calculus Problem with Space Curves and Partial Derivatives **Problem 59:** Let \( P(x, y, z) = y e^{x^2 - z} + \tan^{-1} (xyz) + 3 e^{x + 2} \), and let \( \mathbf{r}(t) \) be a space curve such that \( \mathbf{r}(0) = \left\langle -2, \frac{1}{4}, 4 \right\rangle \) and \( \mathbf{r}'(0) = \langle 1, \beta, -4 \rangle \), where \( \beta \) is some real number. 1. Let \( \alpha \) be a real number such that the rate of change of \( P \) at the point \((-2, -2, 4)\) in the direction of \(\langle \alpha, -1, \alpha \rangle \) is \( \frac{5}{\sqrt{2}} \). 2. Let \( \beta \) be a value such that \(\frac{dP}{dt} = -6\) at \( t = 0 \). Find the value of the product \( \alpha \beta \). --- **Explanation:** This problem involves the application of multivariable calculus, specifically dealing with partial derivatives and the chain rule in the context of space curves. - The function \( P \) is defined in terms of the variables \( x \), \( y \), and \( z \). - The space curve \( \mathbf{r}(t) \) has its position and velocity vectors provided at \( t = 0 \). - To solve this, one typically needs to find the directional derivative of \( P \), determine the necessary value of \( \alpha \) such that a specific rate of change of \( P \) is achieved in a given direction, and then find \( \beta \) so that the overall rate of change of \( P \) with respect to time \( t \) is as specified. Once \( \alpha \) and \( \beta \) are found, their product \( \alpha \beta \) is the final answer.
Transcribed Image Text:### Calculus Problem with Space Curves and Partial Derivatives **Problem 59:** Let \( P(x, y, z) = y e^{x^2 - z} + \tan^{-1} (xyz) + 3 e^{x + 2} \), and let \( \mathbf{r}(t) \) be a space curve such that \( \mathbf{r}(0) = \left\langle -2, \frac{1}{4}, 4 \right\rangle \) and \( \mathbf{r}'(0) = \langle 1, \beta, -4 \rangle \), where \( \beta \) is some real number. 1. Let \( \alpha \) be a real number such that the rate of change of \( P \) at the point \((-2, -2, 4)\) in the direction of \(\langle \alpha, -1, \alpha \rangle \) is \( \frac{5}{\sqrt{2}} \). 2. Let \( \beta \) be a value such that \(\frac{dP}{dt} = -6\) at \( t = 0 \). Find the value of the product \( \alpha \beta \). --- **Explanation:** This problem involves the application of multivariable calculus, specifically dealing with partial derivatives and the chain rule in the context of space curves. - The function \( P \) is defined in terms of the variables \( x \), \( y \), and \( z \). - The space curve \( \mathbf{r}(t) \) has its position and velocity vectors provided at \( t = 0 \). - To solve this, one typically needs to find the directional derivative of \( P \), determine the necessary value of \( \alpha \) such that a specific rate of change of \( P \) is achieved in a given direction, and then find \( \beta \) so that the overall rate of change of \( P \) with respect to time \( t \) is as specified. Once \( \alpha \) and \( \beta \) are found, their product \( \alpha \beta \) is the final answer.
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