- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
14.4 Exercises
- A vector-valued function $F$ satisfies the equation $tF'(t) = F(t) + tA$ for each $t \geq 0,$ where $A$ is a fixed vector. Compute $F''(1)$ and $F(3)$ in terms of $A,$ if $F(1) = 2A..$
- To apply the second fundamental theorem, use the given value for $F(1)$ and calculate $F'(1)$ using the given equation.
- Taking the derivative of the given equation, we get $$tF''(t) = A$$Thus, $F''(1) = A.$ Now, to find $F(3),$ consider that by rearranging the given equation, we can express $F(t)$ as $$F(t) = tF'(t) - tA$$But since $F(1) = 2A,$ this means that $F'(1) = F(1) + A = 3A.$ Rewriting $F''(t) = \frac{A}{t}$ for $t > 0$ with $F'(1) = 3A$ we can integrate $F''(t)$ giving us: $$F'(t) - F'(1) = \int_1^t \frac{A}{x}\,dx$$Evaluating the integral gives us $$F'(t) = A\text{log}\,t + 3A$$Integrating again with $F(1) = 2A,$ we find that $$F(t) = \int_1^t A\text{log}\,x + 3A\,dx + 2A$$Evaluating the integral for $t = 3,$ we get \begin{align*} F(3) &= Ax\text{log}\,x + 2Ax\,\Biggr|_1^3 + 2A \\ &= (6 + 3\text{log}\,3)A \quad \blacksquare\end{align*}