Problem 47 - Calculus Early Transcendentals

Question

Suppose \(w=f(x, y, z)\) and \(\ell\) is the line \(\mathbf{r}(t)=\langle a t, b t, c t\rangle,\) for \(-\infty

Step-by-Step Solution

Answer

In summary, for a given scalar field function \(w = f(x, y, z)\) and a line \(\boldsymbol{r}(t) = \langle at, bt, ct \rangle\), the first derivative of \(w\) with respect to \(t\) is given by: $$w'(t) = w_x a + w_y b + w_z c$$ And the second derivative of \(w\) with respect to \(t\) is given by: $$w''(t) = (\frac{\partial w_x}{\partial x}a + \frac{\partial w_y}{\partial y}a + \frac{\partial w_z}{\partial z}a)a \\ + (\frac{\partial w_x}{\partial x}b + \frac{\partial w_y}{\partial y}b + \frac{\partial w_z}{\partial z}b)b \\ + (\frac{\partial w_x}{\partial x}c + \frac{\partial w_y}{\partial y}c + \frac{\partial w_z}{\partial z}c)c$$ For parts (b) and (c), we found the first derivatives \(w'(t)\) for the given functions \(f(x, y, z) = xyz\) and \(f(x, y, z) = \sqrt{x^2 + y^2 + z^2}\), respectively.

Step 1: Calculate the partial derivatives

Find the partial derivatives of \(w\) with respect to \(x, y, z\): $$w_x = \frac{\partial w}{\partial x},\quad w_y = \frac{\partial w}{\partial y},\quad w_z = \frac{\partial w}{\partial z}$$

Step 2: Apply the Chain Rule

Use the chain rule to find \(w'(t)\): $$w'(t) = w_x \frac{dx}{dt} + w_y \frac{dy}{dt} + w_z \frac{dz}{dt}$$ Since \(\boldsymbol{r}(t) = \langle at, bt, ct \rangle\), we have: $$\frac{dx}{dt} = a,\quad \frac{dy}{dt} = b,\quad \frac{dz}{dt} = c$$ So, the first derivative of \(w\) with respect to \(t\) is: $$w'(t) = w_x a + w_y b + w_z c$$

Step 3: Part A:

The first derivative of \(w\) with respect to \(t\) along the given line \(\boldsymbol{r}(t)\) is: $$w'(t) = w_x a + w_y b + w_z c$$

Step 4: Part B:

Given the function \(f(x, y, z) = xyz\), compute \(w'(t)\): First, calculate the partial derivatives: $$w_x = yz,\quad w_y = xz,\quad w_z = xy$$ Next, calculate the first derivative of \(w\): $$w'(t) = (yz)a + (xz)b + (xy)c$$

Step 5: Part C:

Given the function \(f(x, y, z) = \sqrt{x^2 + y^2 + z^2}\), compute \(w'(t)\): First, calculate the partial derivatives: $$w_x = \frac{x}{\sqrt{x^2 + y^2 + z^2}},\quad w_y = \frac{y}{\sqrt{x^2 + y^2 + z^2}},\quad w_z = \frac{z}{\sqrt{x^2 + y^2 + z^2}}$$ Next, calculate the first derivative of \(w\): $$w'(t) = \frac{x}{\sqrt{x^2 + y^2 + z^2}}a + \frac{y}{\sqrt{x^2 + y^2 + z^2}}b + \frac{z}{\sqrt{x^2 + y^2 + z^2}}c$$

Step 6: Part D:

For a general scalar function \(w = f(x, y, z)\), we will calculate the second derivative, \(w''(t)\).

Step 3: Differentiate \(w'(t)\) with respect to \(t\)

To find \(w''(t)\), we need to differentiate \(w'(t)\), which is: $$w'(t) = w_x a + w_y b + w_z c$$ Taking the derivative with respect to \(t\), we have: $$w''(t) = \frac{d(w_x a)}{dt} + \frac{d(w_y b)}{dt} + \frac{d(w_z c)}{dt}$$ Using chain rule for each term: $$w''(t) = (\frac{\partial w_x}{\partial x}\frac{dx}{dt} + \frac{\partial w_y}{\partial y}\frac{dx}{dt} + \frac{\partial w_z}{\partial z}\frac{dx}{dt})a \\ + (\frac{\partial w_x}{\partial x}\frac{dy}{dt} + \frac{\partial w_y}{\partial y}\frac{dy}{dt} + \frac{\partial w_z}{\partial z}\frac{dy}{dt})b \\ + (\frac{\partial w_x}{\partial x}\frac{dz}{dt} + \frac{\partial w_y}{\partial y}\frac{dz}{dt} + \frac{\partial w_z}{\partial z}\frac{dz}{dt})c$$ Now, since \(\frac{dx}{dt} = a, \frac{dy}{dt} = b,\) and \(\frac{dz}{dt} = c\), we get: $$w''(t) = (\frac{\partial w_x}{\partial x}a + \frac{\partial w_y}{\partial y}a + \frac{\partial w_z}{\partial z}a)a \\ + (\frac{\partial w_x}{\partial x}b + \frac{\partial w_y}{\partial y}b + \frac{\partial w_z}{\partial z}b)b \\ + (\frac{\partial w_x}{\partial x}c + \frac{\partial w_y}{\partial y}c + \frac{\partial w_z}{\partial z}c)c$$ So, for a general scalar function \(w = f(x, y, z)\), the second derivative \(w''(t)\) is given by the above expression.

Key Concepts

Partial Derivatives

Partial derivatives are an essential concept in multivariable calculus. They allow us to find the rate of change of a function with respect to one variable while keeping the other variables constant.
For a function like \( w = f(x, y, z) \), the partial derivatives are written as:\[ w_x = \frac{\partial w}{\partial x}, \quad w_y = \frac{\partial w}{\partial y}, \quad w_z = \frac{\partial w}{\partial z} \] These expressions tell us how the function \( w \) changes as \( x, y, \) and \( z \) change individually. This is useful in fields like physics and engineering where systems depend on multiple variables.

Chain Rule

The chain rule in multivariable calculus helps us differentiate composite functions. When a function depends on multiple variables that are themselves functions of another variable, the chain rule is essential.
If \( w = f(x, y, z) \) and \( \mathbf{r}(t) = \langle at, bt, ct \rangle \), we use the chain rule to find \( w'(t) \) by taking partial derivatives and multiplying by derivatives of the parameterized line:

• \( \frac{dx}{dt} = a \)
• \( \frac{dy}{dt} = b \)
• \( \frac{dz}{dt} = c \)

The result is \( w'(t) = w_x a + w_y b + w_z c \). This allows us to express how \( w \) changes with \( t \), the parameter of the line.

Scalar Functions

Scalar functions, like those found in this exercise, output a single value. In multivariable calculus, these functions can depend on several variables.
Common examples include \( f(x, y, z) = xyz \) or \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \).
Depending on the combination of variables, the function can represent physical quantities like temperature or potential energy. Studying how these functions behave, especially when the variables change, is crucial to understanding complex systems and phenomena.

Calculus of Parametric Curves

When dealing with functions defined along curves, the calculus of parametric curves becomes vital. Parametric equations like \( \mathbf{r}(t) = \langle at, bt, ct \rangle \) describe paths in space and allow us to study changes along these paths.
Calculating derivatives along these paths requires understanding both the direction and magnitude of change.

• \( w'(t) = w_x a + w_y b + w_z c \) for first derivatives.
• Higher-order derivatives, like \( w''(t) \), use further differentiation, exploring how rates of change themselves change.

This exploration within parametric curves provides deeper insight into real-world applications, from mechanics to computer graphics.