If 𝑣=𝑓𝑥,𝑦, define 𝛿𝑣Question 1Answera.𝛿𝑣=∂𝑓∂𝑥𝛿𝑥+∂𝑓∂𝑦𝛿𝑦b.𝛿𝑣=∂𝑣𝛿𝑥+∂𝑣𝛿𝑦c.𝛿𝑣=∂𝑣∂𝑥𝛿𝑥+∂𝑣∂𝑦𝛿𝑦d.𝛿𝑣=∂𝑣∂𝑥+∂𝑣∂𝑦

To define $\delta v$ given $v = f(x, y)$, we will analyze the answer choices provided:

1. Break Down the Problem

We know that $\delta v$ represents the change in $v$ resulting from small changes in $x$ and $y$, denoted as $\delta x$ and $\delta y$. Therefore, we need to examine each option to see which correctly represents the total derivative of $v$.

2. Relevant Concepts

The correct expression for the change in $v$ based on changes in $x$ and $y$ is derived from the multivariable chain rule: $\delta v = \frac{\partial v}{\partial x} \delta x + \frac{\partial v}{\partial y} \delta y$

3. Analysis and Detail

Now let's analyze each option:

1. Option a: $\delta v = \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial y} \delta y$ - This is correct as it uses partial derivatives of $f$ (where $f$ is the function describing $v$).
2. Option b: $\delta v = \frac{\partial v}{\partial x} \delta x + \frac{\partial v}{\partial y} \delta y$ - This is the same formula but incorrect because it uses terms $\partial v$ which is a different variable representation not derived from $f$.
3. Option c: $\delta v = \frac{\partial v}{\partial x} \delta x + \frac{\partial v}{\partial y} \delta y$ - Same issue as option b, so this is incorrect in context.
4. Option d: $\delta v = \frac{\partial v}{\partial x} + \frac{\partial v}{\partial y}$ - This is incorrect as it suggests an additive form without the changes $\delta x$ and $\delta y$.

4. Verify and Summarize

The correct expression for $\delta v$ based on the definition and changes in $x$ and $y$ is given in Option a. Therefore, the analysis shows that Option a correctly follows the mathematical formulation of changes in multivariable functions.

Final Answer

The correct answer is:
a. $\delta v = \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial y} \delta y$