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**majamin** "Slope of velocity" implies acceleration. The vector $\displaystyle \left(\frac{dx}{dt},\frac{dy}{dt}\right)$ is the **velocity vector**, and $\displaystyle \frac{\frac{dx}{dt}}{\frac{dy}{dt}}$ is the **slope of the tangent** of the parametric curve at time t. The magnitude of velocity $\displaystyle |\vec{v}(t)| = \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy} {dt}\right)^2}$. If we want to calculate acceleration, then $\displaystyle \vec{a}(t)=\left(\frac{d^2x}{dt^2},\frac{d^2y}{dt^ 2}\right)$. In general, velocity and acceleration are not orthogonal (perpendicular), but the **unit vector** of velocity and its derivative (the component of acceleration perpendicular to velocity) is. In this case, the product of the slopes would be -1 (true for any perpendicular slopes).