To find $\dfrac{dy}{dt}$ when $x = -3$, we need to use the chain rule, given that $y$ is a function of $x$, and $x$ is a function of $t$.

Step 1: Differentiate $y = 2x^2 - x$ with respect to $t$

Using the chain rule, the derivative of $y$ with respect to $t$ is: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$

Step 2: Find $\frac{dy}{dx}$

First, we need to differentiate $y = 2x^2 - x$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(2x^2 - x) = 4x - 1$

Step 3: Substitute $x = -3$ and $\frac{dx}{dt} = -0.1$

Now, substitute $x = -3$ into $\frac{dy}{dx}$: $\frac{dy}{dx} \bigg|_{x=-3} = 4(-3) - 1 = -12 - 1 = -13$

Step 4: Compute $\frac{dy}{dt}$

Now, substitute $\frac{dy}{dx} = -13$ and $\frac{dx}{dt} = -0.1$ into the chain rule formula: $\frac{dy}{dt} = (-13) \cdot (-0.1) = 1.3$

Final Answer

Therefore, $\dfrac{dy}{dt}$ when $x = -3$ is: $\frac{dy}{dt} = 1.3$

Would you like to explore any of the steps in more detail? Here are some related questions you might be interested in:

1. How does the chain rule work when multiple variables are involved?
2. What would $\dfrac{dy}{dt}$ be if $\dfrac{dx}{dt}$ were a different value?
3. How does the sign of $\dfrac{dx}{dt}$ affect the overall result?
4. Can you find $\dfrac{d^2y}{dt^2}$ for this function?
5. How would you interpret the physical meaning of $\dfrac{dy}{dt}$ in a real-world scenario?

Tip: When working with related rates, always keep track of which variables are dependent on others and apply the chain rule carefully.