Answer :
Certainly! Let's go through the problem step by step to find the directional derivative of the function \(\phi(x, y, z) = x y^2 + y z + x^2\) along the tangent to the curve defined by the parametric equations \(x = t\), \(y = 2t^2\), and \(z = t^3\) at the point \((1, 1, 1)\).
### Step 1: Compute the Gradient of \(\phi\)
First, we need to compute the gradient vector \(\nabla \phi\) of the function \(\phi(x, y, z)\).
[tex]\[ \nabla \phi = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right) \][/tex]
Compute each partial derivative:
[tex]\[ \frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x} (x y^2 + y z + x^2) = y^2 + 2x \][/tex]
[tex]\[ \frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} (x y^2 + y z + x^2) = 2xy + z \][/tex]
[tex]\[ \frac{\partial \phi}{\partial z} = \frac{\partial}{\partial z} (x y^2 + y z + x^2) = y \][/tex]
Thus, the gradient vector is:
[tex]\[ \nabla \phi = (y^2 + 2x, 2xy + z, y) \][/tex]
### Step 2: Evaluate the Gradient at \((1, 1, 1)\)
Next, we need to evaluate this gradient at the point \((1, 1, 1)\):
[tex]\[ \nabla \phi (1, 1, 1) = (1^2 + 2 \cdot 1, 2 \cdot 1 \cdot 1 + 1, 1) = (3, 3, 1) \][/tex]
### Step 3: Compute the Tangent Vector to the Curve at \(t = 1\)
Now we have to find the tangent vector to the curve \(x = t\), \(y = 2t^2\), \(z = t^3\) at \(t = 1\). This involves calculating the derivatives of \(x\), \(y\), and \(z\) with respect to \(t\).
[tex]\[ \frac{dx}{dt} = 1, \quad \frac{dy}{dt} = \frac{d}{dt} (2t^2) = 4t, \quad \frac{dz}{dt} = \frac{d}{dt} (t^3) = 3t^2 \][/tex]
Evaluate these derivatives at \(t = 1\):
[tex]\[ \left. \frac{dx}{dt} \right|_{t=1} = 1 \][/tex]
[tex]\[ \left. \frac{dy}{dt} \right|_{t=1} = 4 \cdot 1 = 4 \][/tex]
[tex]\[ \left. \frac{dz}{dt} \right|_{t=1} = 3 \cdot 1^2 = 3 \][/tex]
Thus, the tangent vector at \(t = 1\) is:
[tex]\[ \mathbf{v} = \left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right) = (1, 4, 3) \][/tex]
### Step 4: Calculate the Directional Derivative
The directional derivative of \(\phi\) in the direction of \(\mathbf{v}\) is given by the dot product of the gradient vector and the unit tangent vector:
[tex]\[ D_{\mathbf{v}} \phi (1, 1, 1) = \nabla \phi (1, 1, 1) \cdot \mathbf{v} \][/tex]
So,
[tex]\[ D_{\mathbf{v}} \phi (1, 1, 1) = (3, 3, 1) \cdot (1, 4, 3) \][/tex]
Calculate the dot product:
[tex]\[ D_{\mathbf{v}} \phi (1, 1, 1) = 3 \cdot 1 + 3 \cdot 4 + 1 \cdot 3 = 3 + 12 + 3 = 18 \][/tex]
### Conclusion
The directional derivative of \(\phi\) at the point \((1, 1, 1)\) along the tangent to the curve is:
[tex]\[ \boxed{18} \][/tex]
### Step 1: Compute the Gradient of \(\phi\)
First, we need to compute the gradient vector \(\nabla \phi\) of the function \(\phi(x, y, z)\).
[tex]\[ \nabla \phi = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right) \][/tex]
Compute each partial derivative:
[tex]\[ \frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x} (x y^2 + y z + x^2) = y^2 + 2x \][/tex]
[tex]\[ \frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} (x y^2 + y z + x^2) = 2xy + z \][/tex]
[tex]\[ \frac{\partial \phi}{\partial z} = \frac{\partial}{\partial z} (x y^2 + y z + x^2) = y \][/tex]
Thus, the gradient vector is:
[tex]\[ \nabla \phi = (y^2 + 2x, 2xy + z, y) \][/tex]
### Step 2: Evaluate the Gradient at \((1, 1, 1)\)
Next, we need to evaluate this gradient at the point \((1, 1, 1)\):
[tex]\[ \nabla \phi (1, 1, 1) = (1^2 + 2 \cdot 1, 2 \cdot 1 \cdot 1 + 1, 1) = (3, 3, 1) \][/tex]
### Step 3: Compute the Tangent Vector to the Curve at \(t = 1\)
Now we have to find the tangent vector to the curve \(x = t\), \(y = 2t^2\), \(z = t^3\) at \(t = 1\). This involves calculating the derivatives of \(x\), \(y\), and \(z\) with respect to \(t\).
[tex]\[ \frac{dx}{dt} = 1, \quad \frac{dy}{dt} = \frac{d}{dt} (2t^2) = 4t, \quad \frac{dz}{dt} = \frac{d}{dt} (t^3) = 3t^2 \][/tex]
Evaluate these derivatives at \(t = 1\):
[tex]\[ \left. \frac{dx}{dt} \right|_{t=1} = 1 \][/tex]
[tex]\[ \left. \frac{dy}{dt} \right|_{t=1} = 4 \cdot 1 = 4 \][/tex]
[tex]\[ \left. \frac{dz}{dt} \right|_{t=1} = 3 \cdot 1^2 = 3 \][/tex]
Thus, the tangent vector at \(t = 1\) is:
[tex]\[ \mathbf{v} = \left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right) = (1, 4, 3) \][/tex]
### Step 4: Calculate the Directional Derivative
The directional derivative of \(\phi\) in the direction of \(\mathbf{v}\) is given by the dot product of the gradient vector and the unit tangent vector:
[tex]\[ D_{\mathbf{v}} \phi (1, 1, 1) = \nabla \phi (1, 1, 1) \cdot \mathbf{v} \][/tex]
So,
[tex]\[ D_{\mathbf{v}} \phi (1, 1, 1) = (3, 3, 1) \cdot (1, 4, 3) \][/tex]
Calculate the dot product:
[tex]\[ D_{\mathbf{v}} \phi (1, 1, 1) = 3 \cdot 1 + 3 \cdot 4 + 1 \cdot 3 = 3 + 12 + 3 = 18 \][/tex]
### Conclusion
The directional derivative of \(\phi\) at the point \((1, 1, 1)\) along the tangent to the curve is:
[tex]\[ \boxed{18} \][/tex]
