Pdf Vector Field Eb91Ae

1. **Problem Statement:** We have a continuous random variable with probability density function (pdf): $$f(t) = \begin{cases} k e^{-t}, & t \geq 0 \\ 0, & \text{otherwise} \end{cases}$$ We need to find: - i. The value of $k$ - ii. The mean (expected value) of the distribution. 2. **Finding the value of $k$:** Since $f(t)$ is a pdf, it must satisfy the normalization condition: $$\int_{-\infty}^{\infty} f(t) \, dt = 1$$ Given $f(t) = 0$ for $t < 0$, the integral reduces to: $$\int_0^{\infty} k e^{-t} \, dt = 1$$ Calculate the integral: $$k \int_0^{\infty} e^{-t} \, dt = k \left[-e^{-t}\right]_0^{\infty} = k (0 - (-1)) = k$$ So, $$k = 1$$ 3. **Finding the mean (expected value) $E[T]$:** The mean is given by: $$E[T] = \int_{-\infty}^{\infty} t f(t) \, dt = \int_0^{\infty} t k e^{-t} \, dt$$ Substitute $k=1$: $$E[T] = \int_0^{\infty} t e^{-t} \, dt$$ Use integration by parts: Let $u = t$, $dv = e^{-t} dt$; Then $du = dt$, $v = -e^{-t}$. So, $$\int t e^{-t} dt = -t e^{-t} + \int e^{-t} dt = -t e^{-t} - e^{-t} + C$$ Evaluate from $0$ to $\infty$: $$\left[-t e^{-t} - e^{-t}\right]_0^{\infty} = (0 + 0) - (0 - 1) = 1$$ Therefore, $$E[T] = 1$$ --- 4. **Vector field problem:** Given vector field: $$\mathbf{F} = x^2 y \mathbf{i} - (z^3 - 3x) \mathbf{j} + 4 y^2 \mathbf{k}$$ Find at point $(1, -2, 1)$: - i. Divergence $\nabla \cdot \mathbf{F}$ - ii. Curl $\nabla \times \mathbf{F}$ 5. **Calculate divergence:** $$\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2 y) + \frac{\partial}{\partial y}[-(z^3 - 3x)] + \frac{\partial}{\partial z}(4 y^2)$$ Calculate each partial derivative: - $\frac{\partial}{\partial x}(x^2 y) = 2 x y$ - $\frac{\partial}{\partial y}[-(z^3 - 3x)] = 0$ (since $z^3$ and $x$ are constants w.r.t. $y$) - $\frac{\partial}{\partial z}(4 y^2) = 0$ (since $y^2$ is constant w.r.t. $z$) So, $$\nabla \cdot \mathbf{F} = 2 x y + 0 + 0 = 2 x y$$ Evaluate at $(1, -2, 1)$: $$2 \times 1 \times (-2) = -4$$ 6. **Calculate curl:** $$\nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2 y & -(z^3 - 3x) & 4 y^2 \end{vmatrix}$$ Calculate components: - $\mathbf{i}$ component: $$\frac{\partial}{\partial y}(4 y^2) - \frac{\partial}{\partial z}[-(z^3 - 3x)] = 8 y - (-3 z^2) = 8 y + 3 z^2$$ - $\mathbf{j}$ component: $$\frac{\partial}{\partial z}(x^2 y) - \frac{\partial}{\partial x}(4 y^2) = 0 - 0 = 0$$ - $\mathbf{k}$ component: $$\frac{\partial}{\partial x}[-(z^3 - 3x)] - \frac{\partial}{\partial y}(x^2 y) = (0 + 3) - x^2 = 3 - x^2$$ Evaluate at $(1, -2, 1)$: - $\mathbf{i}$: $8 \times (-2) + 3 \times 1^2 = -16 + 3 = -13$ - $\mathbf{j}$: $0$ - $\mathbf{k}$: $3 - 1^2 = 3 - 1 = 2$ So, $$\nabla \times \mathbf{F} = -13 \mathbf{i} + 0 \mathbf{j} + 2 \mathbf{k}$$ --- **Final answers:** - i. $k = 1$ - ii. Mean $= 1$ - i. Divergence at $(1, -2, 1)$ is $-4$ - ii. Curl at $(1, -2, 1)$ is $-13 \mathbf{i} + 0 \mathbf{j} + 2 \mathbf{k}$