Find the indicated partial derivative: $u = e^{4r\theta} \sin(\theta)$.

Neetesh Kumar | December 4, 2024

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This is the solution to Math 1D
Assignment: 14.3 Question Number 27
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Step-by-step solution:

Step 1: Compute the first partial derivatives

Partial derivative of $u$ with respect to $r$:

Differentiating $u = e^{4r\theta} \sin(\theta)$ with respect to $r$:

$\frac{\partial u}{\partial r} = \frac{\partial}{\partial r} \left(e^{4r\theta} \sin(\theta)\right)$.

Since $\sin(\theta)$ is treated as constant with respect to $r$:

$\frac{\partial u}{\partial r} = 4\theta e^{4r\theta} \sin(\theta)$.

Step 2: Compute the second partial derivative with respect to $r$:

Differentiating $\frac{\partial u}{\partial r} = 4\theta e^{4r\theta} \sin(\theta)$ again with respect to $r$:

$\frac{\partial^2 u}{\partial r^2} = \frac{\partial}{\partial r} \left(4\theta e^{4r\theta} \sin(\theta)\right)$.

Using the chain rule, we get:

$\frac{\partial^2 u}{\partial r^2} = 16\theta^2 e^{4r\theta} \sin(\theta)$.

Step 3: Compute the third partial derivative with respect to $\theta$:

Differentiating $\frac{\partial^2 u}{\partial r^2} = 16\theta^2 e^{4r\theta} \sin(\theta)$ with respect to $\theta$:

$\frac{\partial^3 u}{\partial r^2 \partial \theta} = \frac{\partial}{\partial \theta} \left(16\theta^2 e^{4r\theta} \sin(\theta)\right)$.

Using the product rule:

$\frac{\partial^3 u}{\partial r^2 \partial \theta} = \frac{\partial}{\partial \theta} \left(16\theta^2 e^{4r\theta}\right) \sin(\theta) + 16\theta^2 e^{4r\theta} \frac{\partial}{\partial \theta} \sin(\theta)$.

Now, calculate each term:

1. $\frac{\partial}{\partial \theta} \left(16\theta^2 e^{4r\theta}\right) = 16 \cdot 2\theta \cdot e^{4r\theta} + 16\theta^2 \cdot 4r \cdot e^{4r\theta} = 32\theta e^{4r\theta} + 64r\theta^2 e^{4r\theta}$.

2. $\frac{\partial}{\partial \theta} \sin(\theta) = \cos(\theta)$.

Substitute back into the equation:

$\frac{\partial^3 u}{\partial r^2 \partial \theta} = \left(32\theta e^{4r\theta} + 64r\theta^2 e^{4r\theta}\right) \sin(\theta) + 16\theta^2 e^{4r\theta} \cos(\theta)$.

Final Answer:

$\frac{\partial^3 u}{\partial r^2 \partial \theta} = \boxed{\left(32\theta e^{4r\theta} + 64r\theta^2 e^{4r\theta}\right) \sin(\theta) + 16\theta^2 e^{4r\theta} \cos(\theta)}$

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