I'm having a hard time understanding where to apply linear approximation. These two problems seem very similar to me, except one does not use linear approximation:
Problem 1: Suppose that you measure the velocity of an object by measuring that it takes $t$ second to travel $x$ meters. The measurement error is $Δx$ meters in distance, and the error in time is $Δt$ seconds. What is the absolute value of the error in the linear approximation for the velocity?
$$v + Δv = \frac{x + Δx}{t + Δt}$$
$$v + Δv = (x+Δx)\cdot(t+Δt)^{-1}$$
$$v + Δv = (x+Δx) \frac {1}{t} (1+\frac{Δt}{t})^{-1}$$
Taking the linear approximation of the right-hand term:
$$v + Δv \approx (x+Δx) \frac {1}{t} (1-\frac{Δt}{t})$$
Then multiplying out:
$$v + Δv ≈ \frac{x}{t} + \frac{Δx}{t} − \frac{x \cdot Δt}{t^{2}} - \frac{Δx \cdot Δt}{t^2}$$
Dropping the last term due to being quadratic:
$$v + Δv ≈ \frac{x}{t} + \frac{Δx}{t} − \frac{x \cdot Δt}{t^{2}}$$
Since $v = \frac{x}{t}$ it can be subtracted from both sides, $$Δv \approx \frac{Δx}{t} − \frac{x \cdot Δt}{t^{2}}$$
Problem 2: Suppose you measure the velocity of a mosquito to be $v$ m/s with an error of $Δv$ m/s. We approximate the mosquito to weigh approximately equal to a grain of rice, which is on the order of magnitude of $m$ kg. What is the error in the kinetic energy? Assume there is no rotational velocity.
$$K = \frac{1}{2}mv^2$$
$$K + ΔK = \frac{1}{2}m(v + Δv)^2$$
$$K + ΔK = \frac{1}{2}m(v^2 + 2vΔv + Δv^2)$$
Dropped the last term due to being quadratic:
$$K + ΔK \approx \frac{1}{2}m(v^2 + 2vΔv)$$
Multiplied out:
$$K + ΔK \approx \frac{1}{2}mv^2 + mvΔv$$
Since $K = \frac{1}{2}mv^2$, subtracted from both sides:
$$ΔK \approx mvΔv$$
Why does the first problem take a linear approximation to help solve for error, but the second doesn't? Is there something I'm missing about what makes a variable special?