Corrected exercises: applications of the derivative (201-SN2-RE)

Step 1, diagram and variables: $x$ = distance from the foot to the wall, $y$ = height of the top, both functions of time. Step 2, relation: by Pythagoras, $x^{2}+y^{2}=25$ (a constant, since the ladder is rigid). Step 3, differentiate with respect to time: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$, so $\frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}$. Step 4, substitute now: $x=3$ gives $y=\sqrt{25-9}=4$, and $\frac{dx}{dt}=0.6$: $\frac{dy}{dt}=-\frac{3}{4}(0.6)=-0.45$ m/s. Step 5, conclusion: the top slides down at $0.45$ m/s, the negative sign signalling a height that DECREASES.

With $x=4.9$: $y=\sqrt{25-24.01}=\sqrt{0.99}\approx 0.995$ m, so $\frac{dy}{dt}=-\frac{4.9}{0.995}(0.6)\approx-2.95$ m/s: more than six times faster than in a). Interpretation: when the ladder is almost flat, $y\to 0$ and the ratio $\frac{x}{y}$ blows up: the speed of the fall tends to infinity even though the foot still moves at 0.6 m/s. The model reaches its limit here (a real ladder would leave the wall), but the mathematical message is right: two quantities linked by a NONLINEAR relation have rates whose ratio keeps changing.

Similar triangles: at any height $h$, the radius of the water surface satisfies $\frac{r}{h}=\frac{3}{6}=\frac{1}{2}$, so $r=\frac{h}{2}$. Volume: $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi\frac{h^{2}}{4}h=\frac{\pi h^{3}}{12}$: a single variable, as required before differentiating. Differentiate: $\frac{dV}{dt}=\frac{\pi h^{2}}{4}\cdot\frac{dh}{dt}$. Substitute: $2=\frac{\pi(16)}{4}\frac{dh}{dt}=4\pi\frac{dh}{dt}$, hence $\frac{dh}{dt}=\frac{1}{2\pi}\approx 0.159$ m/min. Remark: the fuller the cone, the wider the surface and the SLOWER the level rises, at a constant flow rate.

Setting $x=3$ in $x^{2}+y^{2}=25$, he gets $9+y^{2}=25$, so $y=4$: a CONSTANT. Differentiating a constant next, he of course finds $\frac{dy}{dt}=0$. The error is to have frozen a variable BEFORE differentiating: $x=3$ is not a property of the ladder, it is a snapshot, true for a single second. By substituting too early, he turns a motion into a photograph, and a photograph has no speed. Absolute rule: first differentiate the GENERAL relation, valid at every instant, and substitute the numerical values only at the very end.

$f'(x)=3x^{2}-6x=3x(x-2)$, which vanishes at $x=0$ and $x=2$: these are the critical numbers (the derivative exists everywhere, being a polynomial). Signs: for $x<0$, the product of two negatives times 3, so $f'>0$; for $0<x<2$, $f'<0$; for $x>2$, $f'>0$. The function increases, decreases, then increases. By the First Derivative Test: at $x=0$, $f'$ changes from $+$ to $-$: local MAXIMUM, of value $f(0)=2$, that is, the point $(0, 2)$. At $x=2$, $f'$ changes from $-$ to $+$: local MINIMUM, $f(2)=8-12+2=-2$, that is, $(2, -2)$. These are the peak and the trough of the graph.

$f''(x)=6x-6=6(x-1)$, which vanishes at $x=1$ and does change sign: negative for $x<1$ (concave DOWN) and positive for $x>1$ (concave UP). The inflection point is therefore $(1,f(1))=(1, 0)$, where the graph crosses the $x$-axis between its peak and its trough. Reading: to the left of 1 the curve is a dome (it carries its maximum), to the right a bowl (it carries its minimum), and it is exactly at $x=1$ that it switches from one to the other.

Second Derivative Test: $f''(0)=-6<0$, so the curve is concave down there: a critical point in an upside-down bowl is a MAXIMUM ✓. $f''(2)=6>0$, concave up: MINIMUM ✓. Both tests agree. The test fails when $f''(c)=0$: it then concludes NOTHING (it could be a maximum, a minimum or neither), and you must go back to the First Derivative Test, which always works.

$g(x)=x^{4}$: $g'=4x^{3}$, $g''=12x^{2}$. At 0: $g''(0)=0$, the necessary condition holds. Yet $g''=12x^{2}\ge 0$ EVERYWHERE: it vanishes at 0 without ever changing sign, and the curve stays concave up on both sides. There is therefore NO inflection point at 0 (it is in fact a minimum). Lesson: $f''(c)=0$ is a NECESSARY condition but not a SUFFICIENT one; what defines an inflection point is the CHANGE OF SIGN of $f''$, not its vanishing. Exactly as $f'(c)=0$ does not guarantee an extremum ($x^{3}$ at 0): in both cases, only the change of sign counts.

After cutting, the base measures $(60-2x)$ by $(60-2x)$ and the height is $x$: $V(x)=x(60-2x)^{2}$. Realistic domain: we need $x>0$ and $60-2x>0$, so $x\in(0, 30)$. Differentiate (expanding, $V=3600x-240x^{2}+4x^{3}$): $V'=3600-480x+12x^{2}=12(x^{2}-40x+300)=12(x-10)(x-30)$. The roots are $x=10$ and $x=30$; only $x=10$ lies in the open domain. The optimal cut is therefore 10 cm.

First Derivative Test: $V'=12(x-10)(x-30)$. For $0<x<10$, both factors are negative and the product is positive: $V'>0$, increasing. For $10<x<30$, the first factor is positive and the second negative: $V'<0$, decreasing. The derivative changes from $+$ to $-$ at $x=10$: it is indeed a MAXIMUM (and it is absolute on the domain, since the function only rises then falls). Volume: $V(10)=10(60-20)^{2}=10(1600)=16000$ cm³, that is, 16 litres. Check with the second derivative: $V''=24x-480$, and $V''(10)=-240<0$ ✓.

Constraint: $V=\pi r^{2}h=355$, so $h=\frac{355}{\pi r^{2}}$. Quantity to minimize: $A=2\pi r^{2}+2\pi rh$ (the two disks and the side). Substituting: $A(r)=2\pi r^{2}+2\pi r\cdot\frac{355}{\pi r^{2}}=2\pi r^{2}+\frac{710}{r}$. Differentiate: $A'=4\pi r-\frac{710}{r^{2}}$, which vanishes when $4\pi r^{3}=710$, that is, $r=\sqrt[3]{\frac{710}{4\pi}}\approx 3.84$ cm. Then $h=\frac{355}{\pi(3.84)^{2}}\approx 7.67$ cm, and we observe $h\approx 2r$. General proof: from $4\pi r^{3}=2V=2\pi r^{2}h$ we get $h=2r$ directly. The minimum is confirmed by $A''=4\pi+\frac{1420}{r^{3}}>0$ everywhere: the function is concave up, so the critical point is indeed a minimum. Optimal area: $A\approx 277.5$ cm².

Real can: $A=2\pi(3.3)^{2}+2\pi(3.3)(10.4)\approx 68.4+215.7\approx 284.1$ cm², against $277.5$ cm² at the optimum: about $2.4$ % too much metal. Reasons for the gap: first, the GRIP and ergonomics, since a can 7.7 cm tall and 7.7 cm wide would be wide and unstable, awkward to hold and to drink from; next, manufacturing, since the bottom and the lid are thicker than the wall (double seaming, resistance to the gas pressure), so the real cost weighs the disks more heavily and pushes towards a slimmer shape; finally, marketing and logistics, since a tall shape takes less shelf space per litre, stacks better and is recognized from afar. Lesson: the mathematical optimum answers exactly the question asked; if the answer looks off, it is because the real cost function contains terms the model ignored.

$v(t)=s'(t)=3t^{2}-18t+24=3\left(t^{2}-6t+8\right)=3(t-2)(t-4)$ and $a(t)=v'(t)=6t-18=6(t-3)$. The object is at rest when $v(t)=0$, that is, at $t=2$ s and $t=4$ s. The acceleration vanishes at $t=3$ s, which is NOT an instant of rest: it is the instant when the velocity is extremal, here minimal ($v(3)=-3$ m/s).

The sign of $v(t)=3(t-2)(t-4)$: positive for $t<2$, negative for $2<t<4$, positive for $t>4$. The object moves FORWARD on $\left[0, 2\right)$ and on $\left(4, +\infty\right)$; it moves BACKWARD on $\left(2, 4\right)$. The two instants of rest are therefore changes of direction, not mere pauses.

The object speeds up (its SPEED, the absolute value of the velocity, increases) when $v$ and $a$ have the SAME sign, that is, when $v(t)\,a(t)>0$. Let us examine the four zones cut out by $t=2$, $t=3$, $t=4$. On $\left(0, 2\right)$: $v>0$ and $a<0$, a negative product, it SLOWS DOWN (it moves forward while braking towards its stop). On $\left(2, 3\right)$: $v<0$ and $a<0$, a positive product, it SPEEDS UP (it moves backward faster and faster). On $\left(3, 4\right)$: $v<0$ and $a>0$, a negative product, it SLOWS DOWN (it moves backward while braking). On $\left(4, +\infty\right)$: $v>0$ and $a>0$, a positive product, it SPEEDS UP. Conclusion: it speeds up on $\left(2, 3\right)\cup\left(4, +\infty\right)$. The key point is that a NEGATIVE acceleration does not mean slowing down: between $t=2$ and $t=3$, the acceleration is negative and yet the object goes faster and faster, because it is moving in the negative direction. It is the sign of the PRODUCT that decides, never the sign of $a$ alone.

Positions at the key instants: $s(0)=0$, $s(2)=8-36+48=20$, $s(4)=64-144+96=16$, $s(5)=125-225+120=20$. DISPLACEMENT: the net change in position, $s(5)-s(0)=20-0=20$ m. TOTAL DISTANCE: add up the lengths travelled on each interval of constant direction, in absolute value, $\left|20-0\right|+\left|16-20\right|+\left|20-16\right|=20+4+4=28$ m. The two differ because the object TURNS BACK at $t=2$ and at $t=4$: the 4 metres travelled backward cancel 4 metres of the net displacement, but they do add to the distance travelled. This is exactly the distinction between $\int_{0}^{5}v\,dt$ and $\int_{0}^{5}\left|v\right|dt$ that the integral calculus course will formalize. A classic error to avoid: computing $s(5)-s(0)$ and announcing it as the distance travelled, without having looked for the instants of rest.

Let $x$ be the distance from the person to the foot of the lamp and $\ell$ the length of the shadow. Similar triangles (the lamp's and the person's) give $\frac{\ell}{1.8}=\frac{x+\ell}{5}$, that is, $5\ell=1.8x+1.8\ell$, so $3.2\ell=1.8x$ and $\ell=0.5625\,x$. Differentiating with respect to time: $\frac{d\ell}{dt}=0.5625\frac{dx}{dt}=0.5625(1.2)=0.675$ m/s: the shadow gets longer at 0.675 m/s. The TIP of the shadow is at the distance $x+\ell=1.5625\,x$ from the lamp, so it moves at $1.5625(1.2)=1.875$ m/s. The two answers differ, and an exam question does not always say which one is asked: read “the shadow gets longer” (a length) against “the tip of the shadow moves” (a position). Note that both speeds are CONSTANT, independent of $x$: this is special to this configuration of similar triangles.

Let $h$ be the height of the balloon and $\theta$ the angle of elevation. The relation is $\tan\theta=\frac{h}{40}$. Differentiating with respect to time: $\sec^{2}\theta\cdot\frac{d\theta}{dt}=\frac{1}{40}\cdot\frac{dh}{dt}$. At the instant considered, $h=30$ gives $\tan\theta=\frac{30}{40}=0.75$, so $\sec^{2}\theta=1+\tan^{2}\theta=1+0.5625=1.5625$. Isolating: $\frac{d\theta}{dt}=\frac{\frac{1}{40}(3)}{1.5625}=\frac{0.075}{1.5625}=0.048$ rad/s. The angle therefore increases by about 0.048 radian per second, nearly $2.75^{\circ}$ per second. Careful: differentiating requires RADIANS, since $\left(\tan\theta\right)'=\sec^{2}\theta$ only holds in radian measure.

Let $y$ be the distance travelled north, $x$ the distance travelled east, and $z$ the distance between the vehicles. Pythagoras: $z^{2}=x^{2}+y^{2}$. Differentiating: $2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$, so $\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}$. After 2 hours: $x=160$ km, $y=120$ km, and $z=\sqrt{160^{2}+120^{2}}=\sqrt{25600+14400}=\sqrt{40000}=200$ km. Hence $\frac{dz}{dt}=\frac{160(80)+120(60)}{200}=\frac{12800+7200}{200}=\frac{20000}{200}=100$ km/h. Plausibility check: the separation speed, 100 km/h, is exactly $\sqrt{80^{2}+60^{2}}$, which is normal here since both paths are straight and perpendicular: the relative velocity vector is constant.

The five-step method. 1) Draw a DIAGRAM and name every VARYING quantity with a letter, never with its value. 2) Write what is given and what is asked, as derivatives with respect to time. 3) Find a RELATION between the variables (similar triangles, Pythagoras, tangent, volume formula). 4) DIFFERENTIATE this relation with respect to time, applying the chain rule to each variable. 5) ONLY NOW, substitute the numerical values at the instant considered, then isolate the unknown. The fatal error is to swap steps 4 and 5: if you write $h=30$ BEFORE differentiating, then $h$ becomes a constant, its derivative is zero, and all the information about the motion disappears. You would get $\frac{d\theta}{dt}=0$, which is absurd. The rule to remember fits in one sentence: an instantaneous value is valid at one instant only, so it cannot enter a relation you are about to differentiate with respect to time.

Rolle's Theorem: if $f$ is continuous on $[a, b]$, differentiable on $(a, b)$ and if $f(a)=f(b)$, then there is at least one $c\in(a, b)$ such that $f'(c)=0$. Check: $f(x)=x^{2}-4x+3$ is a polynomial, hence continuous and differentiable everywhere; $f(1)=1-4+3=0$ and $f(3)=9-12+3=0$: the values at the endpoints coincide ✓. Then $f'(x)=2x-4$ vanishes at $x=2$, which does belong to $(1, 3)$: the promised $c$ exists, it is the vertex of the parabola.

Mean Value Theorem: the same continuity and differentiability hypotheses, without requiring $f(a)=f(b)$; then there is a $c\in(a, b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$, that is, a tangent line PARALLEL to the chord. Here: $\frac{f(2)-f(0)}{2-0}=\frac{4-0}{2}=2$, the slope of the grey chord. Solve $f'(c)=2c=2$, so $c=1$, the point where the tangent line in the figure touches the curve, and the tangent at $(1, 1)$ is $y=2x-1$, indeed parallel to the chord ✓. Rolle's Theorem is the special case of the Mean Value Theorem where the chord is horizontal.

$f(-1)=|-1|=1$ and $f(1)=1$: the equality at the endpoints holds, and $f$ is continuous on $[-1, 1]$. Yet $f'(x)=-1$ for $x<0$ and $+1$ for $x>0$: the derivative NEVER vanishes. The missing hypothesis is DIFFERENTIABILITY on the whole open interval: at $x=0$, $|x|$ has a corner and is not differentiable. This proves that the differentiability hypothesis is not decorative: a single corner, right in the middle of the interval, is enough for the conclusion of the theorem to collapse.

Let $s(t)$ be the position, continuous and differentiable (a car does not teleport and its velocity exists at every instant). Average speed: $\frac{s(2)-s(0)}{2-0}=\frac{180}{2}=90$ km/h. By the Mean Value Theorem, there is a $c\in(0, 2)$ such that $s'(c)=90$: at that precise instant, the INSTANTANEOUS speed was exactly 90 km/h. The matching device: the AVERAGE SPEED CAMERA (point-to-point speed enforcement), which times a vehicle between two gantries. If the vehicle covered the distance at an average above the limit, the theorem guarantees that it really exceeded that limit at some instant, whatever the driver says: the ticket is mathematically unassailable.

The rule: if $\lim\frac{f}{g}$ has the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, if $f$ and $g$ are differentiable near the point (with $g'\ne 0$), and if $\lim\frac{f'}{g'}$ exists, then $\lim\frac{f}{g}=\lim\frac{f'}{g'}$. First computation: $\frac{\sin x}{x}$ is indeed $\frac{0}{0}$, so differentiate top and bottom: $\frac{\cos x}{1}\to 1$ ✓ (a known result, but beware of circularity: the derivative of sine was established FROM this limit, so L'Hôpital's rule does not prove it, it only recovers it). Second: $\frac{e^{x}-1}{x}$ is $\frac{0}{0}$, so $\frac{e^{x}}{1}\to 1$.

$\frac{1-\cos x}{x^{2}}$: form $\frac{0}{0}$, differentiate: $\frac{\sin x}{2x}$, still $\frac{0}{0}$, differentiate again: $\frac{\cos x}{2}\to\frac{1}{2}$ ✓ (the value found with the conjugate in the chapter on limits). You may apply the rule again as many times as the indeterminate form persists, provided you CHECK at each step that it really does persist. For $\frac{\ln x}{x}$ at $+\infty$: form $\frac{\infty}{\infty}$, differentiate: $\frac{1/x}{1}=\frac{1}{x}\to 0$. A fundamental result: the logarithm grows infinitely more slowly than $x$.

$x\ln x$ has the form $0\cdot(-\infty)$, which the rule does not handle directly: rewrite it as a quotient, $\frac{\ln x}{1/x}$, which is $\frac{-\infty}{\infty}$. Differentiate: $\frac{1/x}{-1/x^{2}}=-x\to 0$. So $\lim_{x\to 0^{+}}x\ln x=0$: this is the limit taken for granted when computing $\int_{0}^{1}\ln x\,dx$ in the chapter on improper integrals. For $\left(1+\frac{1}{x}\right)^{x}$, form $1^{\infty}$: take the logarithm. Let $y$ be the expression; $\ln y=x\ln\left(1+\frac{1}{x}\right)=\frac{\ln(1+1/x)}{1/x}$, of the form $\frac{0}{0}$. Differentiating top and bottom with respect to $x$: $\frac{\frac{-1/x^{2}}{1+1/x}}{-1/x^{2}}=\frac{1}{1+1/x}\to 1$. So $\ln y\to 1$ and $y\to e$: this is the very definition of the number $e$, and the formula for continuously compounded interest.

The error is applying the rule WITHOUT checking for an indeterminate form. At $x=1$: the numerator equals $1+1=2$ and the denominator $1+3=4$: the form is $\frac{2}{4}$, perfectly determinate! Since the function is continuous at 1, the limit is obtained by direct substitution: $\frac{2}{4}=\frac{1}{2}$. The student's result, 2, is therefore wrong by a factor of 4. Lesson: L'Hôpital's rule is NOT a general rule for computing limits, it is a remedy for one precise pathology; applied to a healthy limit, it gives a result with no connection to it. Always evaluate first, and differentiate only if you actually land on $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

The tangent line at the point $(x_{n},f(x_{n}))$ has equation $y=f(x_{n})+f'(x_{n})(x-x_{n})$. We look for where it crosses the axis, so we set $y=0$: $0=f(x_{n})+f'(x_{n})(x-x_{n})$, hence $x-x_{n}=-\frac{f(x_{n})}{f'(x_{n})}$ and finally $x=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$. This abscissa is called $x_{n+1}$: the idea is that a curve looks very much like its tangent line near the point of contact, so where the tangent crosses the axis, the curve almost does too.

Here $f'(x)=2x$, so the formula becomes $x_{n+1}=x_{n}-\frac{x_{n}^{2}-2}{2x_{n}}=\frac{x_{n}}{2}+\frac{1}{x_{n}}$. Start $x_{0}=2$: $x_{1}=1+0.5=1.5$ (the point in the figure). Then $x_{2}=0.75+\frac{1}{1.5}=1.41666667$. Then $x_{3}=0.70833333+\frac{1}{1.41666667}=1.41421569$. Since the exact value is $1.41421356$, three iterations are enough for five decimals.

Correct decimals: $x_{1}=1.5$ gives 0; $x_{2}=1.4166\ldots$ gives 2 (1.41); $x_{3}=1.41421569$ gives 5 (1.41421); $x_{4}$ would give about 11. The number of correct decimals DOUBLES at every step: this is QUADRATIC convergence. Compare with bisection, which gains a factor of 2 per step (about one decimal every 3.3 steps): for ten decimals, Newton needs 4 iterations where bisection requires more than thirty. A useful aside: this recurrence $x_{n+1}=\frac{x_{n}}{2}+\frac{1}{x_{n}}$ was already known to the Babylonians for extracting square roots, more than thirty centuries before Newton.

Failure 1: a derivative that is ZERO or nearly zero at $x_{n}$: the tangent line is horizontal or nearly so, and its intersection with the axis shoots off to infinity; the process diverges. Failure 2: a bad starting point, which can make the sequence oscillate between two values without ever converging, or make it converge to ANOTHER root than the one sought (the classic case of $f(x)=x^{3}-2x+2$ with $x_{0}=0$, which loops forever between 0 and 1). The choice of $x_{0}$ is therefore decisive: the method is only LOCAL, it brilliantly refines an estimate that is already good but does not search globally. In practice the two approaches are combined: a few bisection steps, safe but slow, to get close to the root, then Newton for lightning-fast precision.