Applications and Extensions of Calculus — Standard Exercises

1Exercise Objectives

Problems 1–3 concern applications of single-variable calculus, Problem 4 concerns multivariable calculus, and Problem 5 concerns differential equations. For each problem, identify the theorem or method used and state its conditions of application.

2Problem 1

Determine the intervals of increase and decrease and the local extrema of $f\left(x\right)=x^3-3x$f(x)=x^3-3x.

2.1Sample Solution

The function $f$f is differentiable on all of $\mathbb{R}$\mathbb{R}, and

The critical points are $x=-1$x=-1 and $x=1$x=1. The derivative is positive on $(-\mathrm{\infty },-1)$(-\infty,-1) and $(1,\mathrm{\infty })$(1,\infty) and negative on $(-1,1)$(-1,1). Hence $f$f increases on $(-\mathrm{\infty },-1)$(-\infty,-1) and $(1,\mathrm{\infty })$(1,\infty) and decreases on $(-1,1)$(-1,1). Moreover,

Thus $f$f has a local maximum value $2$2 at $x=-1$x=-1 and a local minimum value $-2$-2 at $x=1$x=1.

2.2Explanation

The equation $f^{\prime }\left(x\right)=0$f'(x)=0 alone does not guarantee an extremum. Check that the sign of the derivative changes across each critical point.

3Problem 2

A particle moves with velocity $v\left(t\right)=t-1$v(t)=t-1 for $0\text{[PARSE ERROR: Undefined("Command("le")")]}t\text{[PARSE ERROR: Undefined("Command("le")")]}3$0\le t\le3. Find its displacement and distance traveled.

3.1Sample Solution

The function $v$v is continuous on $[0,3]$[0,3]. The displacement is

Distance traveled is the integral of speed. Since $v$v changes sign at $t=1$t=1,

3.2Explanation

Displacement is a signed accumulation that retains direction, whereas distance traveled accumulates the speed $\left|v\right|$|v|. Split the interval wherever velocity changes sign.

4Problem 3

Find the average value of $f\left(x\right)=x^2$f(x)=x^2 on $[0,2]$[0,2].

4.1Sample Solution

The function $f$f is continuous on $[0,2]$[0,2], and the interval has positive length. By the definition of average value,

4.2Explanation

A definite integral is the total accumulation over an interval. Divide that total by the interval length to obtain the average value.

5Problem 4

For $f(x,y)=x^{2}y+\sin y$f(x,y)=x^2y+\sin y, compute the partial derivatives $f_x$f_x and $f_y$f_y. Also evaluate

5.1Sample Solution

The function $f$f is continuously differentiable on the entire plane. Treat $y$y as a constant when calculating $f_x$f_x and $x$x as a constant when calculating $f_y$f_y. Thus,

The iterated integral ${\int }_0^1{\int }_0^{2}xydxdy$\int_0^1\int_0^2xy\,dx\,dy means that one first integrates with respect to $x$x while holding $y$y fixed and then integrates the result with respect to $y$y. Because $xy$xy is continuous on the rectangle $[0,2]\times [0,1]$[0,2]\times[0,1], this successive computation agrees with the double integral over that rectangle:

5.2Explanation

A partial derivative fixes the other variable and measures local change in one coordinate direction. A multiple integral accumulates local contributions over a region.

6Problem 5

Let $I$I be a real interval. Find all $C^1$C^1 functions on $I$I satisfying

Include the case $y=0$y=0.

6.1Sample Solution

Let $y$y be any solution on $I$I. Using the integrating factor $e^{-x^2/2}$e^{-x^2/2} gives

Because $I$I is an interval, a function with zero derivative is constant on $I$I. Hence there is some $A\in \mathbb{R}$A\in\mathbb{R} such that

Conversely, every function of this form satisfies

The value $A=0$A=0 gives the zero solution, so this classifies all solutions.

6.2Explanation

The integrating-factor method does not divide by $y$y, so it classifies all solutions without losing the zero solution. If separation of variables is used instead, division by $y$y requires a separate check of the zero solution.

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