From Velocity to Tangent Slope
A planar particle moves along a curve described by its position vector \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\). Its velocity vector is \(\mathbf{v}(t)=\mathbf{r}'(t)=\langle x'(t),y'(t)\rangle\). The components tell us how quickly the particle’s horizontal and vertical coordinates are changing.
At a parameter value \(t_0\), the velocity vector also gives the direction of the curve’s tangent, provided the velocity is not the zero vector. When the horizontal component of velocity is nonzero, the tangent is not vertical, and its slope is the vertical component divided by the horizontal component.
The slope describes the tangent line’s change in \(y\) per unit change in \(x\). The parameter \(t\) is not necessarily time, but in a motion problem it often is. If \(x\) and \(y\) have units of length and \(t\) has units of time, then both velocity components have units of length per time, and their ratio is dimensionless, as a slope should be.
Why the Component Ratio Gives the Slope
The curve is given parametrically: both coordinates depend on \(t\). To find the slope, compare the changes in the two coordinates as the parameter changes. For \(t\ne t_0\), the slope of the secant from \(\mathbf{r}(t_0)\) to \(\mathbf{r}(t)\), when the horizontal change is nonzero, is
As \(t\) approaches \(t_0\), the numerator on the right approaches \(y'(t_0)\), and its denominator approaches \(x'(t_0)\). If \(x'(t_0)\ne0\), division is valid in the limit. The limiting secant slope is therefore \(y'(t_0)/x'(t_0)\). This argument uses the definition of derivative and does not require assuming that \(x\) is one-to-one on a neighborhood of \(t_0\).
Equivalently, if the coordinates can be regarded locally as a differentiable function \(y\) of \(x\), the chain rule gives \(\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\). Solving for \(\frac{dy}{dx}\) when \(\frac{dx}{dt}\ne0\) produces the same formula. The difference-quotient argument makes clear why the ratio describes the tangent slope directly.
Worked Examples
Worked Example: Find the Tangent Slope and Line
Let \(\mathbf{r}(t)=\langle t^2,2t^3\rangle\). Find the tangent slope and tangent line at \(t=1\).
First find the point on the curve:
Thus the point is \((1,2)\). Differentiate each coordinate to find the velocity components:
At \(t=1\), the velocity is \(\mathbf{v}(1)=\langle 2,6\rangle\). Since \(v_x(1)=2\ne0\), the tangent slope is
Using the point-slope form of a line, the tangent line is \(y-2=3(x-1)\), or \(y=3x-1\). The line passes through \((1,2)\), since \(3(1)-1=2\), and its slope is \(3\), as calculated from the velocity components.
Worked Example: Interpret Velocity with Units
A particle’s position, in meters, is \(\mathbf{r}(t)=\langle 3t-1,t^2+1\rangle\), where \(t\) is measured in seconds. Find its tangent slope and tangent line at \(t=2\).
The position at \(t=2\) is
So the particle is at \((5,5)\) meters. Differentiate the coordinate functions:
At \(t=2\), \(\mathbf{v}(2)=\langle 3,4\rangle\) meters per second. The horizontal component is nonzero, so
The tangent line at \((5,5)\) is \(y-5=\frac{4}{3}(x-5)\). Substituting \(x=5\) gives \(y=5\), confirming that the line passes through the particle’s position. Its slope is \(4/3\), matching the velocity-component ratio. The units cancel in that ratio, as expected for slope.
When the Tangent Is Vertical
A tangent line is vertical when its direction has no horizontal component and a nonzero vertical component. For a parametric curve, this happens at \(t_0\) when \(x'(t_0)=0\) and \(y'(t_0)\ne0\). The velocity vector is then \(\langle 0,y'(t_0)\rangle\), which points vertically.
In this case, the ratio \(y'(t_0)/x'(t_0)\) is undefined because its denominator is zero. It is not a finite tangent slope; instead, the tangent line itself is vertical. This geometric conclusion comes from the nonzero velocity vector, not from trying to divide by zero.
Worked Example: Identify a Vertical Tangent
Consider \(\mathbf{r}(t)=\langle t^2,2t\rangle\). Determine the tangent at \(t=0\).
The point on the curve is
Differentiate the coordinate functions:
At \(t=0\), the velocity is \(\mathbf{v}(0)=\langle 0,2\rangle\). Its horizontal component is zero, while its vertical component is nonzero. Therefore the tangent direction is vertical, and the tangent line through \((0,0)\) is \(x=0\).
For \(t\ne0\), the secant slope from \((0,0)\) to \((t^2,2t)\) is
As \(t\) approaches zero, this slope does not approach a finite number; its magnitude grows without bound. That behavior agrees with the vertical tangent found from the velocity vector. Substituting \(x=0\) into the tangent line equation confirms that the line passes through \((0,0)\).
Zero Velocity Does Not Determine the Tangent
If both velocity components are zero, then \(\mathbf{v}(t_0)=\langle 0,0\rangle\). There is no direction in this vector to use as a tangent direction, and the ratio \(y'(t_0)/x'(t_0)\) is \(0/0\), which is undefined. This does not prove that the curve has no tangent. It means only that the velocity components at that parameter value do not determine one through the formula.
Worked Example: Why Zero Velocity Is Inconclusive
Let \(\mathbf{r}(t)=\langle t^2,t^2\rangle\). At \(t=0\), the position and velocity are
The velocity-component ratio at \(t=0\) is \(0/0\), so it gives no slope. But for every parameter value, \(x(t)=t^2=y(t)\). The curve lies on the line \(y=x\), which has slope \(1\). Thus the curve has a tangent there even though the velocity vector at that parameter value is zero. The component-ratio method cannot decide the tangent direction in this case.
A Common Mistake: Dividing by Zero
A tempting but incorrect calculation is to use \(\frac{dy}{dx}=\frac{y'}{x'}\) when \(x'=0\), then call the result “infinite.” Division by zero is undefined, not a numerical operation that produces infinity. Instead, check the velocity components separately. If \(x'=0\) and \(y'\ne0\), the nonzero velocity vector is vertical, so the tangent line is vertical. If both components are zero, this test is inconclusive and requires a different analysis of the curve.
Why This Connection Matters
The velocity vector has both a physical interpretation and a geometric one. Its components describe coordinate rates of change, while the vector itself gives the direction in which the particle is moving at that instant. Their ratio converts those rates into the geometric slope of the curve.
This lets us find a tangent slope without first solving one coordinate equation for \(t\) and substituting into the other. It also helps distinguish a steep but nonvertical tangent from a vertical tangent, and it reminds us to check whether the velocity vector is actually nonzero before using its direction.
Check Your Understanding
Use the position and velocity components carefully in each question.
- If \(x'(t_0)=5\) and \(y'(t_0)=-2\), what is the tangent slope at \(t_0\)?
- A particle is at \((2,3)\) when its velocity is \(\langle 0,-4\rangle\). What is the tangent line?
- Why does \(x'(t_0)=0\) by itself not establish a vertical tangent?
- For \(\mathbf{r}(t)=\langle t^2,3t\rangle\), find the point and tangent direction at \(t=0\).
- If both velocity components are zero at a parameter value, what can and cannot be concluded from the component-ratio formula?
