Module 12: Motion, higher derivatives, and related rates

Motion requires a sign chart

Velocity is $v=2t-4$ m/s and acceleration is $a=2$ m/s$^2$. The particle stops at $t=2$ s. It moves left before $2$ and right afterward. Evaluate $s(0)=0$, $s(2)=-4$, and $s(5)=5$. Displacement is $5-0=\boxed5$ m. Total distance is $|{-4}-0|+|5-(-4)|=4+9=\boxed{13}$ m. Adding absolute displacements over the monotone segments avoids cancellation.

Positive acceleration can reduce speed

Because $v<0$ and $a>0$, velocity is moving toward zero, so the particle is slowing down. Locally, speed is $|v|=-v$. Its derivative is $-a=\boxed{-2\text{ m/s}^2}$. The speed is $6$ m/s at the instant; the negative rate describes how that positive speed changes.

A projectile’s distinct events

The velocity is $y'(t)=20-10t$, so launch velocity is $20$ m/s. At the top, velocity is zero: $t=2$ s. Height there is $40-20=20$ m. Landing satisfies $y(t)=5t(4-t)=0$, giving $t=0$ or $4$; the return-to-ground time is $\boxed4$ s. The physical flight domain is $[0,4]$.

Differentiate area before substituting

Start with $A=\pi r^2$. Differentiating gives $A'=2\pi r r'$. Insert the instantaneous values to obtain $A'=2\pi(3)(0.2)=\boxed{1.2\pi\text{ cm}^2/\text{s}}$. The factor $2\pi r$ has length units, turning a radius rate into an area rate.

Find the missing geometric value first

The ladder equation is $x^2+y^2=169$. At $x=5$, its top height is $y=\sqrt{169-25}=12$ m. Differentiate: $2xx'+2yy'=0$, so

The negative sign means downward. The positive magnitude $5/6$ alone would omit direction.

Use similar-triangle geometry to eliminate radius

Substitute $r=h/2$ into $V=\pi r^2h/3$ to get $V=\pi h^3/12$. Thus $V'=(\pi h^2/4)h'$. At depth $4$, $2=4\pi h'$, giving $\boxed{h'=1/(2\pi)\text{ m/min}}$. The surface area $\pi r^2=4\pi$ m$^2$ at this depth provides an independent check: volume rate divided by surface area gives the same depth rate.

Shadow length and tip position are different variables

Similar triangles give $6/(x+q)=2/q$. Cross-multiply: $6q=2x+2q$, hence $q=x/2$. Differentiating gives $q'=x'/2=\boxed{0.75\text{ m/s}}$. The tip position is $x+q$, so its speed is $x'+q'=1.5+0.75=\boxed{2.25\text{ m/s}}$. Reporting only $q'$ as the tip speed would omit the person’s motion.

Construct an easing polynomial from constraints

The conditions at zero give $d=0$ and $c=0$. At one, $a+b=1$ and $3a+2b=0$. Subtract twice the first equation from the second to get $a=-2$, then $b=3$. Thus $E(u)=3u^2-2u^3$.

On $[0,1]$, $u(1-u)=1/4-(u-1/2)^2\le1/4$, so the maximum speed is $\boxed{3L/(2T)}$ at $t=T/2$. Endpoint velocities vanish, but endpoint accelerations need not; this is not a guarantee of smooth acceleration when stitching arbitrary segments.

Rewrite once, then step each exponent

Rewrite: $s=6t^{-3}+2t^{1/2}$. Differentiate term by term:

Again, with $-\tfrac12-1=\tfrac{-1-2}{2}=-\tfrac32$:

So $a(1)=72-\tfrac12=\boxed{71.5\text{ m/s}^2}$. Final sweep: $-3\to-4\to-5$ and $\tfrac12\to-\tfrac12\to-\tfrac32$, every exponent one step lower than the last.

Two fraction subtractions, then a division by zero

$f'(x)=\tfrac53x^{5/3-1}=\tfrac53x^{2/3}$, because $\tfrac53-1=\tfrac{5-3}{3}=\tfrac23$. Then $f''(x)=\tfrac53\cdot\tfrac23x^{2/3-1}=\tfrac{10}{9}x^{-1/3}$, because $\tfrac23-1=\tfrac{2-3}{3}=-\tfrac13$. To be sure $f'(0)$ exists, use the definition rather than the formula: $\dfrac{f(0+h)-f(0)}{h}=\dfrac{h^{5/3}}{h}=h^{2/3}\to0$, so $f'(0)=0$. (The same check on $x^{1/3}$ gives $h^{-2/3}$, which blows up: a formula’s value is not proof of a derivative.) But $f''(0)$ would need $0^{-1/3}=1/0^{1/3}$, which is undefined, and its difference quotient $\tfrac53h^{2/3}/h=\tfrac53h^{-1/3}$ blows up too: $f''(0)$ does not exist. Near $0$, $f''$ grows without bound in size, positive on the right and negative on the left. The slope changes smoothly through $0$, but the rate at which it changes blows up there, which the graph alone does not show.

One click and one factor of 2 per derivative

$y'=5\cdot(-\sin(2x))\cdot2=-10\sin(2x)$. Then $y''=-10\cos(2x)\cdot2=-20\cos(2x)=-4\cdot5\cos(2x)=\boxed{-4y}$. Then $y'''=-20(-\sin(2x))\cdot2=40\sin(2x)$. The angle stays $2x$ on every line (Angle Fortress), and the Wheel runs $\cos\to-\sin\to-\cos\to\sin$.

Divide by four and keep the remainder

$50=4\cdot12+2$. Twelve full turns of the Wheel change nothing, and two clicks take $\sin\to\cos\to-\sin$. So $\dfrac{\dd^{50}}{\dd x^{50}}\sin x=\boxed{-\sin x}$. For $\sin(2x)$ the clicks are the same, and each of the fifty derivatives also brings out a factor $2$: $\boxed{-2^{50}\sin(2x)}$, about $-1.13\times10^{15}\sin(2x)$.

Workbench the quotient, then substitute

From $2x+2yy'=0$, $y'=-x/y$. Workbench for $-x/y$: $u=-x$, $u'=-1$; $v=y$, $v'=y'$. The quotient rule gives

At $(3,4)$: $y''=\boxed{-\tfrac{25}{64}}$. Check: $y=(25-x^2)^{1/2}$ gives $y'=-x(25-x^2)^{-1/2}$, and the product rule gives $y''=-(25-x^2)^{-1/2}-x^2(25-x^2)^{-3/2}=-25(25-x^2)^{-3/2}$. At $x=3$ that is $-25/16^{3/2}=-25/64$. The two routes agree, and the negative sign says the upper semicircle is concave down.

Two sign rows, then compare them

$v=3t^2-18t+24=3(t-2)(t-4)$ and $a=6t-18=6(t-3)$.

It speeds up on $(2,3)$ and $(4,\infty)$ and slows down on $(0,2)$ and $(3,4)$, the same pattern as Picture 12.1. At $t=3$ it moves left at its greatest speed in that stretch, $|v(3)|=|27-54+24|=3$ m/s.

A diameter is not a radius

The diameter is $50$, so $r=25$ cm. Differentiate the movie: $\dd V/\dd t=4\pi r^2\,\dd r/\dd t$. Freeze the frame: $100=4\pi(625)\,\dd r/\dd t$, so $\dd r/\dd t=\dfrac{100}{2500\pi}=\boxed{\dfrac{1}{25\pi}\approx0.0127\text{ cm/s}}$. Using $r=50$ would give $1/(100\pi)$, four times too small. Units: (cm$^3$/s)/cm$^2$ = cm/s.

Each term keeps its partner

$\dfrac{\dd A}{\dd t}=\dfrac{\dd\ell}{\dd t}\,w+\ell\,\dfrac{\dd w}{\dd t}=(3)(12)+(20)(-2)=36-40=\boxed{-4\text{ cm}^2/\text{s}}$. The area is decreasing at $4$ cm$^2$/s: the shrinking width costs more area than the growing length adds. The dropped-partner product $(3)(-2)=-6$ has the wrong value and the wrong units.

Differentiate, then check with a formula in time

After $10$ s, $x=80$ m east and $y=60$ m north, so $D=\sqrt{80^2+60^2}=100$ m. From $D^2=x^2+y^2$, $2DD'=2xx'+2yy'$, so

Direct route: $D(t)=\sqrt{(8t)^2+(6t)^2}=\sqrt{100t^2}=10t$, so $D'=10$ m/s at every moment. Both routes agree. Because both drones started together, their separation grows at a constant rate.

Differentiate the tangent, then freeze at $45^\circ$

With $h$ in metres, $\tan\theta=h/2000$. Differentiate with respect to time (Magic Triangles): $\sec^2\theta\cdot\dfrac{\dd\theta}{\dd t}=\dfrac{1}{2000}\dfrac{\dd h}{\dd t}$. At $h=2000$, $\tan\theta=1$, so $\theta=\pi/4$ and $\sec^2\theta=1+\tan^2\theta=2$. Then

Units: (m/s)/m = 1/s, and an angle rate in calculus is in radians per second.

A product of a falling factor and a rising factor

(a) Nesting dolls inside $\rho$: $\dfrac{\dd\rho}{\dd t}=\rho_0e^{-h/H}\cdot\left(-\dfrac1H\dfrac{\dd h}{\dd t}\right)=-\dfrac{\rho}{H}\dfrac{\dd h}{\dd t}$. Then the product rule on $q=\tfrac12\rho\cdot v^2$:

(b) $h/H=11/8.5\approx1.2941$, so $\rho=1.225e^{-1.2941}\approx1.225(0.27414)\approx0.3358$ kg/m$^3$. Then $q=\tfrac12(0.3358)(400)^2\approx\boxed{26.9\text{ kPa}}$. The bracket is $\dfrac{2(12)}{400}-\dfrac{280}{8500}=0.0600-0.0329=0.0271$ per second, so $\dd q/\dd t\approx26{,}865(0.0271)\approx\boxed{727\text{ Pa/s}}$, about $0.73$ kPa/s. Yes, $q$ is still rising.

(c) Set the bracket to zero: $\dfrac{\dd v}{\dd t}=\dfrac{v}{2H}\dfrac{\dd h}{\dd t}=\dfrac{400(280)}{2(8500)}\approx\boxed{6.6\text{ m/s}^2}$. Less thrust means less acceleration, so the rising-speed term shrinks and $q$ levels off at a lower peak: the structure never sees the load it would have seen at full thrust. In the straight-line model of part (d) the peak load is $q^*=\rho_0e^{-1}aH/\sin\gamma$, directly proportional to $a$.

(d) Here $\dd v/\dd t=a$ and $v=at$, so $\dfrac2v\dfrac{\dd v}{\dd t}=\dfrac2t$; and $\dfrac{\dd h}{\dd t}=at\sin\gamma$. Setting the bracket to zero gives $\dfrac2t=\dfrac{at\sin\gamma}{H}$, so $t^2=\dfrac{2H}{a\sin\gamma}$. Then