Category: Apostol – Calculus 1

Solutions to Calculus Problems

Find a vector valued function satisfying given properties

by RoRi

June 1, 2016

If $A$ and $B$ are fixed vectors and $F$ is a vector valued function with

$F''(t) = tA + B,$

find $F(t)$ if $F(0) = D$ and $F'(0) = C$ .

We have

$\begin{align*} && F''(t) &= tA + B \\ \implies && \int F''(t) \,dt &= \int (tA+B) \, dt \\ \implies && F'(t) &= \frac{1}{2}t^2 A + tB + C \\ \implies && F(t) &= \frac{1}{6} t^3 A + \frac{1}{2}t^2 B + tC + D. \end{align*}$

Prove the zero derivative theorem for vector valued functions

by RoRi

June 1, 2016

Prove the zero derivative theorem for vector valued functions. In other words, show that if $F'(t) = 0$ for each $t$ in an open interval $I$ , then there exists a vector $C$ such that $F(t) = C$ for all $t \in I$ .

Proof. If $F'(t) = 0$ then we have

$\begin{align*} && (f_1'(t), \ldots, f_n'(t)) &= (0, \ldots, 0) \\ \implies && f_i'(t) &= 0 & \text{for all } i. \end{align*}$

By the zero derivative for real valued functions we then have $f_i (t) = c_i$ for some constant $c_i$ for each $i = 1, \ldots, n$ . Therefore we have

$(f_1(t), \ldots, f_n (t)) = (c_1, \ldots, c_n) \quad \implies \quad F(t) = C. \qquad \blacksquare$

Prove a condition for a vector valued function to be differentiable

by RoRi

June 1, 2016

If $F$ is a vector valued function, prove that $F$ is differentiable on the open interval $I$ if and only if for all $t \in I$ we have

$F'(t) = \lim_{h \to 0} \frac{1}{h} (F(t+h) - F(t)).$

Proof. First, assume $F$ is differentiable on $I$ . Then, by the definition of differentiability for vector valued functions we have

$F'(t) &= (f_1'(t), \ldots, f_n'(t))$

and each of the $f_i'(t)$ exists. Therefore by the usual definition of the derivative we have

$\begin{align*} F'(t) &= \left( \lim_{h \to 0} \frac{f_1 (t+h) - f_1(t)}{h}, \ldots, \lim_{h \to 0} \frac{f_n (t+h) - f_n (t)}{h} \right) \\[9pt] &= \lim_{h \to 0} \frac{1}{h} \left( (f_1 (t+h), \ldots, f_n (t+h)) - (f_1 (t), \ldots, f_n (t)) \right) \\[9pt] &= \lim_{h \to 0} \frac{1}{h} (F(t+h) - F(t)). \end{align*}$

Conversely, assume that

$F'(t) = \lim_{h \to 0} \frac{1}{h} (F(t+h) - F(t)).$

Then, the limit

$\lim_{h \to 0} \frac{1}{h} (f_i (t+h) - f_i(t))$

exists for each $i = 1, \ldots, n$ . Thus, $f_i'(t)$ exists for each $i$ . Therefore, $F'(t)$ exists $. \qquad \blacksquare$

Prove a property of limits of vector valued functions

by RoRi

June 1, 2016

Prove that

$\lim_{t \to p} F(t) = A \iff \lim_{t \to p} \lVert F(t) - A \rVert = 0.$

Proof. Let

$F(t) = (f_1(t), \ldots, f_n(t)), \qquad \text{and} \qquad A = (a_1, \ldots, a_n).$

Then,

$\begin{align*} \lim_{t \to p} F(t) = A && \iff && \lim_{t \to p} (f_1(t), \ldots, f_n (t)) &= (a_1, \ldots, a_n) \\ && \iff && \left( \lim_{t \to p} f_1(t), \ldots, \lim_{t \to p} f_n (t) \right) &= (a_1, \ldots, a_n). \end{align*}$

This holds if and only if

$\lim_{t \to p} f_i (t) = a_i \qquad \text{for } i = 1, \ldots, n.$

From the usual limit this is true if and only if

$\lim_{t \to p} | f_i (t) - a_i | = 0 \qquad \text{for } i = 1, \ldots, n.$

This is if and only if

$\begin{align*} &&\left( \lim_{t \to p} |f_1 (t) - a|, \ldots, \lim_{t \to p} |f_n (t) - a| \right) &= (0, \ldots, 0) \\ \iff && \lim_{t \to p} |F(t) - A| = 0. \qquad \blacksquare \end{align*}$

Compute G′ if G = F F′ x F′′

by RoRi

June 1, 2016

Let $G$ be the scalar triple product $F \cdot F' \times F''$ . Prove that $G' = F \cdot F' \times F'''$ .

Proof. We compute,

$\begin{align*} G = F \cdot F' \times F'' && \implies && G' &= (F \cdot F' \times F'')' \\ && \implies && G' &= F' \cdot (F' \times F'') + F \cdot (F' \times F'')' \\ && \implies && G' &= 0 + F \cdot (F'' \times F'' + F' \times F''') \\ && \implies && G' &= F \cdot F' \times F'''. \qquad \blacksquare \end{align*}$

Compute G′ if G = F x F′

by RoRi

June 1, 2016

Let $G = F \times F'$ . Compute $G'$ in terms of $F$ and $F'$ .

Using the product rule for the cross product we compute,

$\begin{align*} G = F \times F' && \implies && G' &= (F \times F')' \\ && \implies && G' &= F' \times F' + F \times F'' \\ && \implies && G' &= F \times F'' \end{align*}$

Since $F' \times F' = 0$ since a vector crossed with itself is 0.

Prove that F′′(t) has the same direction as a given F(t)

by RoRi

June 1, 2016

Let $A, B$ be nonzero vectors and

$F(t) = e^{2t} A + e^{-2t} B.$

Prove that $F''(t)$ has the same direction as $F(t)$ .

Proof. To show that $F(t)$ and $F''(t)$ have the same direction we must show that there exists a constant $c$ such that $F(t) = cF''(t)$ . So, we compute,

$\begin{align*} F(t) = e^{2t}A + e^{-2t}B && \implies && F'(t) &= 2e^{2t}A - 2 e^{-2t} B \\ && \implies && F''(t) &= 4 e^{2t} A + 4e^{-2t} B \\ && \implies && F(t) &= \frac{1}{4} F''(t). \qquad \blacksquare \end{align*}$

Prove that F”(t) is orthogonal to F'(t) under given conditions

by RoRi

June 1, 2016

Let $B$ be a nonzero vector and $F$ a vector valued function with $F(t) \cdot B = t$ for all $t$ , and such that the angle between $F'(t)$ and $B$ is constant. Prove that $F''(t)$ and $F'(t)$ are orthogonal.

Proof. Since $F(t) \cdot B = t$ we have $F'(t) \cdot B = c$ for some constant $c$ . Since the angle between $F'(t)$ and $B$ is constant we have

$\frac{F'(t) \cdot B}{\lVert F'(t) \rVert \lVert B \rVert} = d$

for some constant $d$ . Therefore, $\lVert F'(t) \rVert$ is constant. Hence, $\lVert F'(t) \rVert^2$ is constant, say $a$ . So, we have

$\begin{align*} && \lVert F'(t) \rVert^2 &= a \\ \implies && F'(t) \cdot F'(t) &= a \\ \implies && ( F'(t) \cdot F'(t)) &= 0 \\ \implies && 2F'(t) \cdot F''(t) &= 0 \\ \implies && F'(t) \cdot F''(t) &= 0. \qquad \blacksquare \end{align*}$

Compute the dot product of a vector and a vector valued integral

by RoRi

June 1, 2016

Let $A = 2 \mathbf{i} - 4 \mathbf{j} + \mathbf{k}$ and let

$B = \int_0^1 (te^{2t} \mathbf{i} + t \cosh (2t) \mathbf{j} + 2te^{2t} \mathbf{k}) \, dt.$

Compute the dot product $A \cdot B$ .

First, we evaluate the vector valued integral,

$\begin{align*} B &= \int_0^1 (te^{2t} \mathbf{i} + t \cosh (2t) \mathbf{j} + 2te^{2t} \mathbf{k}) \, dt \\[9pt] &= \left( \frac{1}{4} + \frac{1}{4}e^2, \frac{1}{4} + \frac{1}{2} \sinh 2 - \frac{1}{4} \cosh 2, \frac{1}{2} - \frac{3}{2} e^{-2} \right). \end{align*}$

Therefore, the dot product is

$\begin{align*} A \cdot B &= \frac{1}{2} + \frac{1}{2} e^2 - 1 - 2 \sinh 2 + \cosh 2 + \frac{1}{2} - \frac{3}{2} e^{-2} \\[9pt] &= \frac{1}{2}e^2 - e^2 + e^{-2} + \frac{1}{2}e^2 + \frac{1}{2} e^{-2} - \frac{3}{2} e^{-2} \\[9pt] &= 0. \end{align*}$

Compute a vector valued integral

by RoRi

June 1, 2016

Compute the vector valued integral

$\int_0^1 (te^t \mathbf{i} + t^2 e^t \mathbf{j} + te^{-t} \mathbf{k}) \, dt.$

We compute

$\begin{align*} \int_0^1 (te^t \mathbf{i} + t^2 e^t \mathbf{j} + te^{-t} \mathbf{k} ) \, dt &= \left( \int_0^1 te^t \, dt, \int_0^1 t^2 e^t \, dt, \int_0^1 te^{-t} \, dt \right) \\[9pt] &= \left( 1, e-2, 1 - \frac{2}{e} \right). \end{align*}$