Chapter 12: Problem 20
In Exercises \(19-22,\) functions \(z=f(x, y), x=g(s, t)\) and \(y=h(s, t)\) are given. (a) Use the Multivariable Chain Rule to compute \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\) (b) Evaluate \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\) at the indicated \(s\) and \(t\) values.$ 12. $$ z=\cos \left(\pi x+\frac{\pi}{2} y\right), \quad x=s t^{2}, \quad y=s^{2} t ; \quad s=1, t=1 $$
Short Answer
Step by step solution
Identify the Functions and Variables
Apply Multivariable Chain Rule for \(\frac{\partial z}{\partial s}\)
Apply Multivariable Chain Rule for \(\frac{\partial z}{\partial t}\)
Simplify and Substitute \(s = 1\) and \(t = 1\)
Evaluate \(\sin\left(\frac{3\pi}{2}\right)\)
Calculate \(\frac{\partial z}{\partial t}\) with Substitution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Multivariable Chain Rule
- \( z = \cos(\pi x + \frac{\pi}{2} y) \)
- \( x = st^2 \)
- \( y = s^2 t \)
- \(\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s}\)
Similarly, for \(\frac{\partial z}{\partial t}\):
- \(\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t}\)
Partial Derivatives
- \(\frac{\partial z}{\partial x} = -\sin(\pi x + \frac{\pi}{2} y) \cdot \pi\)
- \(\frac{\partial z}{\partial y} = -\sin(\pi x + \frac{\pi}{2} y) \cdot \frac{\pi}{2} \)
Partial derivatives provide vital insights when functions have multiple inputs, giving rise to more powerful analysis and applications, like optimization and the monitoring of rates of change.
Function Evaluation
- For \(s = 1\) and \(t = 1\), evaluate \(x\) and \(y\):
- \(x = 1\cdot 1^2 = 1\)
- \(y = 1^2 \cdot 1 = 1\)
- \(\frac{\partial z}{\partial s} = 2\pi\)
- \(\frac{\partial z}{\partial t} = 2\pi + \frac{\pi}{2}\)