Questions: Obtenha dF/dt pela regra da cadeia, sendo F a função composta de f, com x e y nos seguintes casos: c) f(x, y)=ln(x^2+y^2), x(t)=3t e y(t)=t-1

Solution Steps

To find \(\frac{dF}{dt}\) using the chain rule, we need to:

1. Compute the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).
2. Compute the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
3. Use the chain rule formula: \(\frac{dF}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}\).

We need to find the derivative of \( F \) with respect to \( t \) using the chain rule. The function \( F \) is a composition of \( f \), where \( f(x, y) = \ln(x^2 + y^2) \), \( x(t) = 3t \), and \( y(t) = t - 1 \).

First, we express \( F \) as a function of \( t \): \[ F(t) = f(x(t), y(t)) = \ln((3t)^2 + (t - 1)^2) \]

To find \(\frac{dF}{dt}\), we use the chain rule: \[ \frac{dF}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} \]

First, compute the partial derivatives of \( f \): \[ f(x, y) = \ln(x^2 + y^2) \] \[ \frac{\partial f}{\partial x} = \frac{1}{x^2 + y^2} \cdot 2x = \frac{2x}{x^2 + y^2} \] \[ \frac{\partial f}{\partial y} = \frac{1}{x^2 + y^2} \cdot 2y = \frac{2y}{x^2 + y^2} \]

Next, compute the derivatives of \( x(t) \) and \( y(t) \): \[ x(t) = 3t \] \[ \frac{dx}{dt} = 3 \] \[ y(t) = t - 1 \] \[ \frac{dy}{dt} = 1 \]

Substitute \( x(t) \) and \( y(t) \) into the partial derivatives: \[ \frac{\partial f}{\partial x} = \frac{2(3t)}{(3t)^2 + (t - 1)^2} = \frac{6t}{9t^2 + t^2 - 2t + 1} = \frac{6t}{10t^2 - 2t + 1} \] \[ \frac{\partial f}{\partial y} = \frac{2(t - 1)}{(3t)^2 + (t - 1)^2} = \frac{2(t - 1)}{10t^2 - 2t + 1} \]

Combine the results using the chain rule: \[ \frac{dF}{dt} = \left( \frac{6t}{10t^2 - 2t + 1} \right) \cdot 3 + \left( \frac{2(t - 1)}{10t^2 - 2t + 1} \right) \cdot 1 \] \[ \frac{dF}{dt} = \frac{18t}{10t^2 - 2t + 1} + \frac{2(t - 1)}{10t^2 - 2t + 1} \] \[ \frac{dF}{dt} = \frac{18t + 2(t - 1)}{10t^2 - 2t + 1} \] \[ \frac{dF}{dt} = \frac{18t + 2t - 2}{10t^2 - 2t + 1} \] \[ \frac{dF}{dt} = \frac{20t - 2}{10t^2 - 2t + 1} \]

Final Answer