Balloon and sphere

✍️ Worked examples

• Example 1:
• A spherical balloon is inflated so that its volume grows at $8\pi$ cm${}^3$/s.
• Find the rate of increase of the radius and of the surface area when $r=4$ cm.
• Solution:
• $V={\displaystyle \frac{4}{3}}\pi r^3$, so ${\displaystyle \frac{dV}{dt}}=4\pi r^2{\displaystyle \frac{dr}{dt}}$.
• Substitute 8π=4π(16)drdt, hence drdt=18 cm/s.
• $A=4\pi r^2$, so ${\displaystyle \frac{dA}{dt}}=8\pi r{\displaystyle \frac{dr}{dt}}=8\pi (4)\left({\displaystyle \frac{1}{8}}\right)=4\pi$ cm${}^2$/s.

Growing sand pile

✍️ Worked examples

• Example 2 Sand falls at $18\pi$ cm${}^3$/s forming a cone whose height always equals its base radius. How fast is the radius increasing when $r=3$ cm?
• Solution:
• With h=r V=13πr2h=13πr3.
• ${\displaystyle \frac{dV}{dt}}=\pi r^2{\displaystyle \frac{dr}{dt}}$.
• $18\pi =\pi (9){\displaystyle \frac{dr}{dt}}$, so ${\displaystyle \frac{dr}{dt}}=2$ cm/s.

Sliding ladder

✍️ Worked examples

• Example 3:
• A 250 cm ladder leans against a vertical wall.
• Its top slides down at 10 cm/s.
• When the foot is 70 cm from the wall find (a) the speed of the foot, (b) the rate of change of the angle between the ladder and the ground.
• Solution:
• Let $x$ be the foot’s distance from the wall and $y$ the height of the top: $x^2+y^2=250^2$.
• When x=70 y=62500−4900=240 cm.
• Differentiate 2xdxdt+2ydydt=0, so dxdt=−yxdydt.
• With dydt=−10 dxdt=−24070(−10)=2407≈34.3 cm/s. The foot slides away from the wall.
• (b) $\sin \theta ={\displaystyle \frac{y}{250}}$, so $\cos \theta {\displaystyle \frac{d\theta }{dt}}={\displaystyle \frac{1}{250}}{\displaystyle \frac{dy}{dt}}$.
• $\cos \theta ={\displaystyle \frac{70}{250}}$, hence ${\displaystyle \frac{70}{250}}{\displaystyle \frac{d\theta }{dt}}={\displaystyle \frac{-10}{250}}$ and ${\displaystyle \frac{d\theta }{dt}}=-{\displaystyle \frac{1}{7}}$ rad/s.
• The angle is decreasing at ${\displaystyle \frac{1}{7}}$ rad/s.

Triangle with changing legs

✍️ Worked examples

• Example 4:
• The legs of a right triangle are 12 cm and 16 cm.
• The first leg grows at 2 cm/s and the second shrinks at 1 cm/s.
• Find the rate of change of the area after 2 s and the time when the triangle becomes isosceles.
• Solution:
• After $t$ seconds the legs are $x=12+2t$ and $y=16-t$.
• $A={\displaystyle \frac{1}{2}}xy$, so ${\displaystyle \frac{dA}{dt}}={\displaystyle \frac{1}{2}}(y{\displaystyle \frac{dx}{dt}}+x{\displaystyle \frac{dy}{dt}})={\displaystyle \frac{1}{2}}[2(16-t)-(12+2t)]=10-2t$.
• At t=2 dAdt=6 cm2/s.
• Isosceles when x=y 12+2t=16−t, so t=43 s.

Rectangular box

✍️ Worked examples

• Example 5:
• A cuboid has a square base of side $x$ that increases at 1 cm/min, and a height $h$ that decreases at 2 cm/min.
• When $x=5$ cm and $h=20$ cm find the rate of change of the volume, and the time after which the volume stops increasing.
• Solution:
• $V=x^{2}h$, so ${\displaystyle \frac{dV}{dt}}=2xh{\displaystyle \frac{dx}{dt}}+x^2{\displaystyle \frac{dh}{dt}}$.
• Now 2(5)(20)(1)+25(−2)=200−50=150 cm3/min.
• The volume stops increasing when ${\displaystyle \frac{dV}{dt}}=0$, i.e. $2xh=2x^2$, so $h=x$.
• Here $x=5+t$ and $h=20-2t$, so $5+t=20-2t$ and $t=5$ min.

Shadow problem

✍️ Worked examples

• Example 6:
• A man 1.8 m tall walks toward a lamp post 5.4 m high at 1.2 m/s.
• Find the rate of change of the length of his shadow, and the rate of change of the distance from his head to the lamp when he is 4.8 m from the post.
• Solution:
• Let $x$ be his distance from the post and $y$ the length of his shadow.
• Similar triangles 5.41.8=x+yy, so 3y=x+y and y=x2.
• ${\displaystyle \frac{dy}{dt}}={\displaystyle \frac{1}{2}}{\displaystyle \frac{dx}{dt}}={\displaystyle \frac{1}{2}}(-1.2)=-0.6$ m/s: the shadow shortens at 0.6 m/s.
• The vertical gap between lamp and head is $5.4-1.8=3.6$ m, so the head-to-lamp distance satisfies $M^2=x^2+3.6^2$.
• When $x=4.8$: $M=\sqrt{23.04+12.96}=6$ m.
• $2M{\displaystyle \frac{dM}{dt}}=2x{\displaystyle \frac{dx}{dt}}$, so ${\displaystyle \frac{dM}{dt}}={\displaystyle \frac{4.8(-1.2)}{6}}=-0.96$ m/s.

Triangle with a changing angle

✍️ Worked examples

• Example 7:
• Two sides $a$ and $b$ of a triangle grow at 0.1 cm/s each, and the angle $\theta$ between them grows at 0.05 rad/s.
• Find the rate of change of the area when $a=b=10$ cm and $\theta ={\displaystyle \frac{\pi }{3}}$.
• Solution:
• $A={\displaystyle \frac{1}{2}}ab\sin \theta$.
• ${\displaystyle \frac{dA}{dt}}={\displaystyle \frac{1}{2}}[({\displaystyle \frac{da}{dt}}b+a{\displaystyle \frac{db}{dt}})\sin \theta +ab\cos \theta {\displaystyle \frac{d\theta }{dt}}]$.
• $={\displaystyle \frac{1}{2}}[(1+1){\displaystyle \frac{\sqrt{3}}{2}}+100\cdot {\displaystyle \frac{1}{2}}\cdot 0.05]={\displaystyle \frac{1}{2}}[\sqrt{3}+2.5]\approx 2.12$ cm${}^2$/s.

Observer and rising balloon

✍️ Worked examples

• Example 8:
• A balloon rises vertically from point $A$.
• An observer at $B$, 200 m from $A$ on the ground, sees the elevation angle $\theta$ increase at 0.12 rad/min.
• How fast is the balloon rising when $\theta ={\displaystyle \frac{\pi }{4}}$?
• Solution:
• $h=200\tan \theta$, so ${\displaystyle \frac{dh}{dt}}=200{\sec }^2\theta {\displaystyle \frac{d\theta }{dt}}$.
• At $\theta ={\displaystyle \frac{\pi }{4}}$, ${\sec }^2\theta =2$.
• ${\displaystyle \frac{dh}{dt}}=200(2)(0.12)=48$ m/min.