In the following question:(a) Express dw

dt as a function of t, both by using the Chain Rule and

by expressing w in terms of t and differentiating directly with respect to t. (b) Then evaluate

dw

dt at the given value of t.

w = ln(x

2 + y

2 + z

2

), x = cost, y = sin t, z = 42t t = 3

We are given the following:

- \( w = \ln(x^2 + y^2 + z^2) \),
- \( x = \cos(t), y = \sin(t), z = 42t \),
- \( t = 3 \).

We need to:

(a) **Express \( \frac{dw}{dt} \) as a function of \( t \)** using both the Chain Rule and by expressing \( w \) directly in terms of \( t \), then differentiating.

(b) **Evaluate \( \frac{dw}{dt} \) at \( t = 3 \)**.

---

### (a) Using the Chain Rule

First, we apply the Chain Rule to differentiate \( w \) with respect to \( t \). Since \( w \) is a function of \( x, y, z \), and each of these is a function of \( t \), we use:

\[
\frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} + \frac{\partial w}{\partial z} \cdot \frac{dz}{dt}
\]

First, let's compute the partial derivatives of \( w \) with respect to \( x, y, z \):

\[
\frac{\partial w}{\partial x} = \frac{2x}{x^2 + y^2 + z^2}, \quad \frac{\partial w}{\partial y} = \frac{2y}{x^2 + y^2 + z^2}, \quad \frac{\partial w}{\partial z} = \frac{2z}{x^2 + y^2 + z^2}
\]

Next, we compute the derivatives of \( x, y, z \) with respect to \( t \):

\[
\frac{dx}{dt} = -\sin(t), \quad \frac{dy}{dt} = \cos(t), \quad \frac{dz}{dt} = 42
\]

Thus, the total derivative becomes:

\[
\frac{dw}{dt} = \frac{2x}{x^2 + y^2 + z^2} \cdot (-\sin t) + \frac{2y}{x^2 + y^2 + z^2} \cdot (\cos t) + \frac{2z}{x^2 + y^2 + z^2} \cdot (42)
\]

Factoring out \( \frac{2}{x^2 + y^2 + z^2} \), we get:

\[
\frac{dw}{dt} = \frac{2}{x^2 + y^2 + z^2} \left( x(-\sin t) + y(\cos t) + z(42) \right)
\]

Now substitute \( x = \cos t, y = \sin t, z = 42t \) into the expression:

\[
\frac{dw}{dt} = \frac{2}{(\cos^2 t + \sin^2 t + (42t)^2)} \left( \cos t (-\sin t) + \sin t (\cos t) + 42t (42) \right)
\]

Simplify:

\[
\frac{dw}{dt} = \frac{2}{1 + (42t)^2} \left( 0 + 42t \cdot 42 \right)
\]

\[
\frac{dw}{dt} = \frac{2 \cdot 42^2 t}{1 + 42^2 t^2}
\]

Thus:

\[
\frac{dw}{dt} = \frac{3528 t}{1 + 1764 t^2}
\]

---

### (b) Evaluating \( \frac{dw}{dt} \) at \( t = 3 \)

Substitute \( t = 3 \) into the expression:

\[
\frac{dw}{dt} = \frac{3528 \cdot 3}{1 + 1764 \cdot 3^2}
\]

Now, let's calculate this expression:

\[
\frac{dw}{dt} = \frac{3528 \cdot 3}{1 + 1764 \cdot 9} = \frac{10584}{1 + 15876} = \frac{10584}{15877}
\]

Thus:

\[
\frac{dw}{dt} \approx 0.6665
\]

---

Do you want to go into more detail on any step, or clarify anything?

Here are some related questions:

1. What are the conditions for applying the Chain Rule in multiple variables?
2. How would the process change if the functions \( x(t), y(t), z(t) \) had different forms?
3. What is the significance of logarithmic differentiation in solving similar problems?
4. How do partial derivatives differ from total derivatives in this context?
5. What are other applications of the Chain Rule in physics or engineering?

**Tip:** Always check the domain of the function after differentiating, especially when logarithmic terms are involved.

In the following question:(a) Express dw/dt as a function of t, both by using the Chain Rule and by expressing w in terms of t and differentiating directly with respect to t. (b) Then evaluate dw/dt at the given value of t. w = ln(x^2 + y^2 + z^2), x = cost, y = sin t, z = 42t, t = 3

Explore the process of using the Chain Rule to differentiate the function w = ln(x^2 + y^2 + z^2) with respect to t. Includes step-by-step solutions and the evaluation of dw/dt at t = 3. Suitable for students studying college-level calculus.

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