# Is the relation "slope=velocity" mathematically valid?

$$\text{Slope= tan(angle with respect to positive X-axis)= scalar output}$$

$$\text{velocity= a vector }$$ Source: Hugh D Young_ Roger A Freedman - University Physics with Modern Physics In SI Units (2019, Pearson) Page-67

Then, doubts are;

1.Is the relation "Slope of tangent=instantaneous x-velocity" valid, as it would mean "scalar=vector"?

2.Even if I write "Slope of tangent=instantaneous x-speed", if the tangent makes obtuse angle slope will be negative, and we know that instantaneous speed is magnitude of instantaneous velocity, which makes instantaneous speed a positive term. So, what exactly does the slope give?

Extra information: Source: Hugh D Young_ Roger A Freedman - University Physics with Modern Physics In SI Units (2019, Pearson) Page-67

Similar confusion arises in,$$\text{ "Area under a x-t graph = change in x-velocity from time 0 to time t"}$$, with right hand side vector and left hand side (area) as scalar. I think answer to the original question, provides solution to this confusion as well.

• 3blue1brown has a series of calculus videos that answers conceptual questions like this. The first is The Essence of Calculus, Chapter 1 Feb 11 at 1:26

Keep in mind there is a difference between a vector and a component of a vector. The velocity is a vector, the x-component of the velocity (or the "$$x$$-velocity" in the language your book uses) is just a component of a vector, which is a scalar.
The velocity vector can be expanded in terms of unit vectors $$\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$$ (which satisfy $$\mathbf{e}_x\cdot \mathbf{e}_x=1, \mathbf{e}_x\cdot \mathbf{e}_y=0$$, etc): $$\begin{equation} \mathbf{v}=v_x \mathbf{e}_x + v_y \mathbf{e}_y + v_z \mathbf{e}_z \end{equation}$$ The slope of the line you wrote down gives you $$v_x$$, which is a scalar given by $$v_x=\mathbf{v}\cdot \mathbf{e}_x$$. This quantity is a scalar, so there's no problem setting it equal to a slope. It is also a signed quantity, so it's not a speed (there's no requirement that $$v_x>0$$).
• @DavidWhite $v_x=\mathbf{v}\cdot{e}_x$ is a scalar, not a vector, and this is the quantity I mean by "the $x$ component of $\mathbf{v}$" (I also think this is standard usage). Of course $v_x \mathbf{e}_x$ is a vector but I would not refer to this as the $x$ component of $\mathbf{v}$. Feb 11 at 1:38