📝 Unit 3: Derivative Applications

Related Rates · Optimization · Motion · Marginal Analysis

🔄 Not Graded — Unlimited Retakes

Purpose: Self-check applied derivative problem-solving.

Score: 0 / 12

Topic 3.1 — Motion

Question 1

Position $s(t)=t^3-9t^2+24t$ (m, s). Find velocity at $t=2$.

Solution:

$v(t)=3t^2-18t+24$; $v(2)=12-36+24=0$ m/s.

Question 2

Same s(t): when is the particle at rest?

$t=2$ and $t=4$
$t=0$ only
$t=3$
Never

Solution:

$3(t^2-6t+8)=3(t-2)(t-4)=0$ → $t=2,4$.

Question 3

Acceleration of $s(t)=t^3-9t^2+24t$ at $t=3$ is:

Solution:

$a(t)=6t-18$; $a(3)=0$. The particle changes from decelerating to accelerating.

Topic 3.2 — Related Rates

Question 4

A balloon's radius grows at 2 cm/s. How fast is the volume changing when $r=5$ cm? ($V=\frac{4}{3}\pi r^3$)

Solution:

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=4\pi(25)(2)=200\pi\approx 628.32$ cm³/s.

Question 5

A 5 m ladder slides down a wall. The base moves out at 1 m/s. When the base is 3 m from the wall, how fast is the top sliding down?

Solution:

$x^2+y^2=25$; when $x=3$, $y=4$. $2x x'+2y y'=0$ → $y'=-x x'/y=-3(1)/4=-0.75$ m/s. Speed = 0.75 m/s down.

Question 6

A conical tank (apex down) has height = 2(radius). Water flows in at 10 m³/min. How fast is the water level rising when depth is 4 m? ($V=\frac{1}{3}\pi r^2 h$)

Solution:

$r=h/2$, so $V=\frac{\pi h^3}{12}$. $\frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}$. $10=\frac{\pi(16)}{4}\frac{dh}{dt}$ → $\frac{dh}{dt}=\frac{10}{4\pi}\approx 0.796$ m/min.

Topic 3.3 — Optimization

Question 7

Two numbers sum to 20. Maximize their product.

Solution:

$P=x(20-x)=20x-x^2$; $P'=20-2x=0$ → $x=10$. Max $P=10\cdot10=100$.

Question 8

A box (open top, square base) has volume 32 m³. Find dimensions minimizing surface area. State the side of the base.

Solution:

Let base $=x$, height $=h$. $x^2 h=32$ → $h=32/x^2$. $S=x^2+4xh=x^2+128/x$. $S'=2x-128/x^2=0$ → $x^3=64$ → $x=4$ m (height = 2 m).

Question 9

Find the point on $y=x^2$ closest to $(6,3)$. Give the x-coordinate.

Solution:

Minimize $D^2=(x-6)^2+(x^2-3)^2$. $(D^2)'=2(x-6)+2(x^2-3)(2x)=2x-12+4x^3-12x=4x^3-10x-12$. Solve $4x^3-10x-12=0$ → $x=1.5$? Let's verify: $4(3.375)-15-12=13.5-27=-13.5$. Solving numerically gives $x\approx 1$. At $x=1$: $P'=4-10-12=-18$. The exact root is between 1.5 and 2; closest x ≈ 1.85.

Topic 3.4 — Marginal Analysis

Question 10

Cost $C(x)=0.01x^2+5x+200$ ($, units). Find marginal cost $C'(50)$.

Solution:

$C'(x)=0.02x+5$; $C'(50)=1+5=$6$/unit.

Question 11

Revenue $R(x)=20x-0.05x^2$. Find production level maximizing revenue.

Solution:

$R'=20-0.1x=0$ → $x=200$ units.

Question 12

A particle's velocity is $v(t)=t^2-4t$. When is it speeding up? (Hint: $v$ and $a$ same sign.)

\(0
\(2
$t>4$
b and c

Solution:

$a=2t-4$. On $t>4$: $v>0,\,a>0$ → speeding up. On \(0

📊 Self-Reflection

Rate confidence (1–4):

Motion (velocity, acceleration)

Related rates

Optimization

Marginal analysis