Calculus with Vector Functions — Question 8

Question 8

Let $$\mathbf F(t)=\int_1^{t^2}\left\langle e^u,\ \frac 1u,\ \cos(\pi u)\right\rangle\,du, \qquad t>0.$$ Tasks

1. Find $\mathbf F'(t)$ without first evaluating the integral.

2. Evaluate $\mathbf F'(1)$ and $\mathbf F(1)$.

3. Explain the roles of the Fundamental Theorem of Calculus and the chain rule.

Question 8 – Solution

Strategy. Differentiate the vector integral componentwise. The moving upper limit contributes its derivative by the chain rule.

Step 1: General rule If $$\mathbf F(t)=\int_a^{g(t)}\mathbf f(u)\,du,$$ then the Fundamental Theorem gives the derivative with respect to the upper limit, while the chain rule multiplies by $g'(t)$: $$\mathbf F'(t)=\mathbf f(g(t))g'(t).$$

Step 2: Apply the rule Here $g(t)=t^2$, so $g'(t)=2t$. Therefore $$\begin{align*} \mathbf F'(t) &=2t\left\langle e^{t^2},\ \frac 1{t^2},\ \cos(\pi t^2)\right\rangle\\ &=\boxed{\left\langle 2te^{t^2},\ \frac 2t,\ 2t\cos(\pi t^2)\right\rangle}. \end{align*}$$ The restriction $t>0$ ensures the interval between $1$ and $t^2$ never crosses the singularity $u=0$.

Step 3: Values at $t=1$ $$\boxed{\mathbf F'(1)=\left\langle 2e,2,-2\right\rangle}.$$ Also, an integral with identical limits is the zero vector: $$\boxed{\mathbf F(1)=\int_1^1\mathbf f(u)\,du=\left\langle 0,0,0\right\rangle}.$$ These results distinguish accumulated value from its instantaneous rate of change.