Tangent Vector Computations at Specified Parameters

Computing a Tangent Vector at a Chosen Parameter

A parametrized space curve assigns a position vector to each allowed parameter value. Its derivative records the curve’s instantaneous direction and rate of change. When a parameter value is specified, the calculation has two distinct parts: differentiate the component functions, then evaluate those derivatives at that parameter.

Keeping those parts separate helps prevent common errors. Substituting the parameter into the original curve gives a point on the curve; substituting it into the derivative gives the tangent vector there. These are different vectors with different roles.

Definition: Let \(\mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle\) be a vector-valued function whose component functions are differentiable at \(t=a\). The tangent vector at parameter \(a\) is \(\mathbf{r}'(a)=\langle f'(a),g'(a),h'(a)\rangle\).

The parameter \(a\) must be in the domain of the curve, and the component derivatives must exist there. A nonzero tangent vector gives a direction along the curve at that parameter. If \(\mathbf{r}'(a)=\mathbf{0}\), the derivative exists but does not give a nonzero tangent direction at that parameter.

A Reliable Calculation Procedure

For a curve in three-dimensional space, differentiate the first, second, and third components independently. Then substitute the specified parameter into each derivative. If useful, also calculate \(\mathbf{r}(a)\) to identify the point where the tangent vector applies—but do not confuse that position with the derivative.

Write the components clearly.
Identify each scalar function in the vector-valued curve.

Differentiate component by component.
Use the appropriate scalar differentiation rules on each component.

Substitute the specified parameter.
Evaluate every component of the derivative at the same parameter value.

Check the result.
Review the derivative rules and substitutions, and distinguish \(\mathbf{r}'(a)\) from the curve point \(\mathbf{r}(a)\).

Why the Derivative Is Computed Componentwise

The componentwise procedure follows directly from the definition of a vector derivative. For \(\mathbf{r}(t)=\langle f(t),g(t),h(t)\rangle\), subtract the vectors at \(t\) and \(a\), divide by the scalar \(t-a\), and take the limit as \(t\) approaches \(a\). Vector limits exist exactly componentwise, so when all three component derivatives exist, the resulting limit has those derivatives as its components.

\begin{aligned} \mathbf{r}'(a) &=\lim_{t\to a}\frac{\mathbf{r}(t)-\mathbf{r}(a)}{t-a}\\ &=\lim_{t\to a} \left\langle \frac{f(t)-f(a)}{t-a}, \frac{g(t)-g(a)}{t-a}, \frac{h(t)-h(a)}{t-a} \right\rangle\\ &=\left\langle \lim_{t\to a}\frac{f(t)-f(a)}{t-a}, \lim_{t\to a}\frac{g(t)-g(a)}{t-a}, \lim_{t\to a}\frac{h(t)-h(a)}{t-a} \right\rangle\\ &=\langle f'(a),g'(a),h'(a)\rangle. \end{aligned}

This argument also explains the order of the practical calculation. Each scalar derivative is computed first, and the parameter is evaluated afterward. The componentwise result requires each component derivative to exist at \(a\); if even one does not exist, this calculation does not establish the vector derivative there.

Worked Examples

Worked Example: A Polynomial Curve

Find the tangent vector to \(\mathbf{r}(t)=\langle t^3-2t,\;2t^2+1,\;5-t\rangle\) at \(t=2\). Also find the point on the curve at that parameter.

First differentiate each component. For the first component, the power rule gives \(\frac{d}{dt}(t^3-2t)=3t^2-2\). For the second, \(\frac{d}{dt}(2t^2+1)=4t\). For the third, \(\frac{d}{dt}(5-t)=-1\). Thus

\mathbf{r}'(t)=\langle 3t^2-2,\;4t,\;-1\rangle.

Now substitute \(t=2\) into the derivative, component by component:

\begin{aligned} \mathbf{r}'(2) &=\langle 3(2)^2-2,\;4(2),\;-1\rangle\\ &=\langle 3(4)-2,\;8,\;-1\rangle\\ &=\langle 12-2,\;8,\;-1\rangle\\ &=\langle 10,\;8,\;-1\rangle. \end{aligned}

The curve point is found by substituting \(t=2\) into the original components, not the derivative:

\begin{aligned} \mathbf{r}(2) &=\langle 2^3-2(2),\;2(2)^2+1,\;5-2\rangle\\ &=\langle 8-4,\;2(4)+1,\;3\rangle\\ &=\langle 4,\;9,\;3\rangle. \end{aligned}

Therefore, the curve passes through \(\langle 4,9,3\rangle\) at \(t=2\), and its tangent vector there is \(\langle 10,8,-1\rangle\). As a check, differentiating the displayed components of \(\mathbf{r}'(t)\) is not needed to obtain the answer; instead, verify the original component derivatives: the derivatives of \(t^3-2t\), \(2t^2+1\), and \(5-t\) are respectively \(3t^2-2\), \(4t\), and \(-1\). Evaluating these at \(2\) gives \(10\), \(8\), and \(-1\), as reported.

Worked Example: A Trigonometric Curve

Find the tangent vector to \(\mathbf{r}(t)=\langle\sin(2t),\;\cos t,\;\sin t+\cos t\rangle\) at \(t=\frac{\pi}{2}\), and identify the curve point there.

Differentiate the first component using the chain rule: the derivative of \(\sin(2t)\) is \(\cos(2t)\cdot 2=2\cos(2t)\). The derivatives of \(\cos t\), \(\sin t\), and \(\cos t\) are \(-\sin t\), \(\cos t\), and \(-\sin t\), respectively. Therefore

\mathbf{r}'(t)=\langle 2\cos(2t),\;-\sin t,\;\cos t-\sin t\rangle.

Substitute \(t=\frac{\pi}{2}\). The angles needed are \(2t=\pi\) and \(t=\frac{\pi}{2}\), with \(\cos\pi=-1\), \(\sin\frac{\pi}{2}=1\), and \(\cos\frac{\pi}{2}=0\):

\begin{aligned} \mathbf{r}'\left(\frac{\pi}{2}\right) &=\left\langle 2\cos\pi,\;-\sin\frac{\pi}{2},\;\cos\frac{\pi}{2}-\sin\frac{\pi}{2} \right\rangle\\ &=\langle 2(-1),\;-1,\;0-1\rangle\\ &=\langle -2,\;-1,\;-1\rangle. \end{aligned}

For the point on the curve, substitute into \(\mathbf{r}(t)\):

\begin{aligned} \mathbf{r}\left(\frac{\pi}{2}\right) &=\left\langle \sin\pi,\;\cos\frac{\pi}{2},\;\sin\frac{\pi}{2}+\cos\frac{\pi}{2} \right\rangle\\ &=\langle 0,\;0,\;1+0\rangle\\ &=\langle 0,\;0,\;1\rangle. \end{aligned}

The tangent vector at the point \(\langle 0,0,1\rangle\) is \(\langle-2,-1,-1\rangle\). To check the calculation, differentiate each original component: \(2\cos(2t)\) follows from the chain rule, \(-\sin t\) is the derivative of \(\cos t\), and \(\cos t-\sin t\) is the derivative of \(\sin t+\cos t\). The exact trigonometric values above then give the stated vector.

Worked Example: Mixed Components with a Product and a Chain Rule

Find \(\mathbf{r}'\left(\frac{\pi}{2}\right)\) for \(\mathbf{r}(t)=\langle t\sin t,\;t^2+\cos(2t),\;e^{-t}\rangle\), and calculate the curve point at the same parameter.

The first component is a product, so use the product rule: the derivative of \(t\sin t\) is \(1\cdot\sin t+t\cdot\cos t=\sin t+t\cos t\). In the second component, the derivative of \(t^2\) is \(2t\), and the chain rule gives \(\frac{d}{dt}\cos(2t)=-2\sin(2t)\). For the third component, the chain rule gives \(\frac{d}{dt}e^{-t}=-e^{-t}\). Hence

\mathbf{r}'(t)=\langle \sin t+t\cos t,\;2t-2\sin(2t),\;-e^{-t}\rangle.

Evaluate all three components at \(t=\frac{\pi}{2}\):

\begin{aligned} \mathbf{r}'\left(\frac{\pi}{2}\right) &=\left\langle \sin\frac{\pi}{2}+\frac{\pi}{2}\cos\frac{\pi}{2}, \;2\left(\frac{\pi}{2}\right)-2\sin\pi, \;-e^{-\pi/2} \right\rangle\\ &=\left\langle 1+\frac{\pi}{2}(0), \;\pi-2(0), \;-e^{-\pi/2} \right\rangle\\ &=\langle 1,\;\pi,\;-e^{-\pi/2}\rangle. \end{aligned}

The point itself is

\begin{aligned} \mathbf{r}\left(\frac{\pi}{2}\right) &=\left\langle \frac{\pi}{2}\sin\frac{\pi}{2}, \;\left(\frac{\pi}{2}\right)^2+\cos\pi, \;e^{-\pi/2} \right\rangle\\ &=\left\langle \frac{\pi}{2}(1), \;\frac{\pi^2}{4}+(-1), \;e^{-\pi/2} \right\rangle\\ &=\left\langle \frac{\pi}{2},\;\frac{\pi^2}{4}-1,\;e^{-\pi/2} \right\rangle. \end{aligned}

Thus the tangent vector is \(\langle1,\pi,-e^{-\pi/2}\rangle\). Check the mixed rules directly: differentiating \(t\sin t\) gives \(\sin t+t\cos t\); differentiating \(t^2+\cos(2t)\) gives \(2t-2\sin(2t)\); and differentiating \(e^{-t}\) gives \(-e^{-t}\). At \(\frac{\pi}{2}\), the values \(\sin\frac{\pi}{2}=1\), \(\cos\frac{\pi}{2}=0\), and \(\sin\pi=0\) produce exactly the three components displayed.

Common Pitfall: Substituting Before Differentiating

A frequent error is to substitute the specified parameter into the curve first and then treat the resulting point as the tangent vector. For example, in the polynomial example, \(\mathbf{r}(2)=\langle4,9,3\rangle\). That is the curve’s position at \(t=2\), not its tangent vector. The tangent vector comes from the derivative:

\mathbf{r}'(t)=\langle3t^2-2,\;4t,\;-1\rangle \quad\Longrightarrow\quad \mathbf{r}'(2)=\langle10,8,-1\rangle.

Another error is to omit the inner derivative in a trigonometric component. For instance, writing \(\frac{d}{dt}\sin(2t)=\cos(2t)\) misses the factor \(2\). The chain rule gives \(\frac{d}{dt}\sin(2t)=2\cos(2t)\), so at \(t=\frac{\pi}{2}\) that component is \(2\cos\pi=-2\), not \(\cos\pi=-1\).

Takeaway: Differentiate first, then substitute. Keep the curve point \(\mathbf{r}(a)\) separate from the tangent vector \(\mathbf{r}'(a)\), and apply product and chain rules before evaluating.

When a Tangent Vector Is Zero

The componentwise calculation can produce the zero vector. For example, if \(\mathbf{r}(t)=\langle t^2,t^2,t^2\rangle\), then \(\mathbf{r}'(t)=\langle2t,2t,2t\rangle\), so \(\mathbf{r}'(0)=\langle0,0,0\rangle\). This is the derivative at the specified parameter, but it does not provide a nonzero direction vector there. Therefore, do not assume that every parameter value yields a nonzero tangent vector. Check the evaluated components before interpreting the result as a direction.

These calculations are useful because they turn a curve’s component formulas into a concrete local direction at a chosen parameter. The essential habits are consistent across polynomial, trigonometric, and mixed-component curves: differentiate each component correctly, evaluate all components at the same parameter, and verify whether the result is distinct from the position vector and nonzero.

Check Your Understanding

For each question, keep the derivative calculation separate from evaluation at the specified parameter.

For \(\mathbf{r}(t)=\langle t^2,\;3t-1,\;t^3\rangle\), find \(\mathbf{r}'(1)\) and \(\mathbf{r}(1)\).
For \(\mathbf{r}(t)=\langle\cos(3t),\;\sin t,\;2\cos t\rangle\), find the derivative formula. Which chain-rule factor appears in the first component?
For \(\mathbf{r}(t)=\langle t\cos t,\;t^2+\sin(2t),\;e^t\rangle\), find \(\mathbf{r}'(0)\).
State the component differentiability condition needed to compute \(\mathbf{r}'(a)\) using the componentwise formula.
What does it mean if the computed tangent vector at a parameter is the zero vector?