Consider the following. \( w=x y^{2}+x^{2} z+y z^{2}, \quad x=t^{2}, \quad y=2 t, \quad z=2 \) (a) Find \( d w / d t \) using the appropriate Chain Rule. \( \frac{d w}{d t}=\square \) (b) Find \( d w / d t \) by converting \( w \) to a function of \( t \) before differentiating. \( \frac{d w}{d t}=\square \)

We are given \[ w = xy^2 + x^2z + yz^2,\quad x = t^2,\quad y = 2t,\quad z = 2. \] ### (a) Using the Chain Rule We start with the chain rule: \[ \frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}. \] **Step 1. Compute the partial derivatives of \( w \).** 1. The partial derivative with respect to \( x \): \[ \frac{\partial w}{\partial x} = \frac{\partial}{\partial x}\left(xy^2\right) + \frac{\partial}{\partial x}\left(x^2z\right) + \frac{\partial}{\partial x}\left(yz^2\right) = y^2 + 2xz + 0 = y^2 + 2xz. \] 2. The partial derivative with respect to \( y \): \[ \frac{\partial w}{\partial y} = \frac{\partial}{\partial y}\left(xy^2\right) + \frac{\partial}{\partial y}\left(x^2z\right) + \frac{\partial}{\partial y}\left(yz^2\right) = 2xy + 0 + z^2 = 2xy + z^2. \] 3. The partial derivative with respect to \( z \): \[ \frac{\partial w}{\partial z} = \frac{\partial}{\partial z}\left(xy^2\right) + \frac{\partial}{\partial z}\left(x^2z\right) + \frac{\partial}{\partial z}\left(yz^2\right) = 0 + x^2 + 2yz = x^2 + 2yz. \] **Step 2. Compute the derivatives of \( x \), \( y \), and \( z \) with respect to \( t \).** - \( x = t^2 \) gives \[ \frac{dx}{dt} = 2t. \] - \( y = 2t \) gives \[ \frac{dy}{dt} = 2. \] - \( z = 2 \) gives \[ \frac{dz}{dt} = 0. \] **Step 3. Substitute into the chain rule formula.** \[ \frac{dw}{dt} = \left(y^2+2xz\right)(2t) + \left(2xy+z^2\right)(2) + \left(x^2+2yz\right)(0). \] Since the last term vanishes, we have: \[ \frac{dw}{dt} = \left(y^2+2xz\right)(2t) + \left(2xy+z^2\right)(2). \] **Step 4. Substitute the given values \( x=t^2,\, y=2t,\, z=2 \).** - Calculate \( y^2 \): \[ (2t)^2 = 4t^2. \] - Calculate \( 2xz \): \[ 2(t^2)(2) = 4t^2. \] - So, \( y^2+2xz = 4t^2+4t^2 = 8t^2 \). - Next, calculate \( 2xy \): \[ 2(t^2)(2t) = 4t^3. \] - And \( z^2 \): \[ 2^2 = 4. \] - So, \( 2xy+z^2 = 4t^3 + 4 \). Now, substitute back: \[ \frac{dw}{dt} = \left(8t^2\right)(2t) + \left(4t^3+4\right)(2) = 16t^3 + 8t^3 + 8. \] \[ \frac{dw}{dt} = 24t^3 + 8. \] Thus, for part (a), \[ \frac{dw}{dt} = 24t^3+8. \] ### (b) Converting \( w \) to a Function of \( t \) First **Step 1. Substitute \( x = t^2 \), \( y = 2t \), and \( z = 2 \) into \( w \).** \[ w = (t^2)(2t)^2 + (t^2)^2(2) + (2t)(2)^2. \] **Step 2. Simplify each term.** 1. First term: \[ (t^2)(2t)^2 = t^2(4t^2) = 4t^4. \] 2. Second term: \[ (t^2)^2(2) = t^4\cdot 2 = 2t^4. \] 3. Third term: \[ (2t)(2)^2 = 2t\cdot 4 = 8t. \] Combine them: \[ w = 4t^4 + 2t^4 + 8t = 6t^4 + 8t. \] **Step 3. Differentiate \( w(t) \) with respect to \( t \).** \[ \frac{dw}{dt} = \frac{d}{dt}(6t^4 + 8t) = 24t^3 + 8. \] Thus, for part (b), \[ \frac{dw}{dt} = 24t^3+8. \]