Example 1 – Derivative as a Rate Measure

Practice Questions with Step by Step Solutions

Question 1

A particle moves so that its position is [s(t)=3t^{3}-9t^{2}+6t] (metres). Find (a) velocity [\dfrac{ds}{dt}] and (b) acceleration [\dfrac{d^{2}s}{dt^{2}}] at [t=2\ \text{s}].

Step-by-step Solution:

1. Compute velocity: [\dfrac{ds}{dt}][=\dfrac{d}{dt}(3t^{3}-9t^{2}+6t)][=9t^{2}-18t+6].
2. Evaluate at [t=2]: [9(4)-18(2)+6][=36-36+6][=6\ \text{m/s}].
3. Compute acceleration: [\dfrac{d^{2}s}{dt^{2}}][=\dfrac{d}{dt}(9t^{2}-18t+6)][=18t-18].
4. Evaluate at [t=2]: [18(2)-18][=36-18][=18\ \text{m/s}^{2}].

Conclusion: [\dfrac{ds}{dt}\big|_{2}][=6\ \text{m/s},][\quad][ \dfrac{d^{2}s}{dt^{2}}\big|_{2}][=18\ \text{m/s}^{2}.]

Question 2

The radius of a circular ripple increases at the rate [\dfrac{dr}{dt}=0.1\ \text{m/s}]. Find the rate of change of the circumference when [r=5\ \text{m}]. (Circumference [C=2\pi r].)

Step-by-step Solution:

1. Differentiate circumference: [\dfrac{dC}{dt}][=\dfrac{d}{dt}(2\pi r)][=2\pi\dfrac{dr}{dt}].
2. Substitute: [\dfrac{dC}{dt}][=2\pi(0.1)][=0.2\pi\ \text{m/s}].
3. Numeric approx (optional): [0.2\pi\approx 0.628\ \text{m/s}].

Conclusion: [\dfrac{dC}{dt}][=0.2\pi\ \text{m/s}] when [r=5\ \text{m}] (note: depends only on [dr/dt]).

Question 3

Water flows into a vertical cylindrical tank of radius [4\ \text{m}] at the rate [\dfrac{dV}{dt}=10\ \text{m}^{3}/\text{s}]. Find the rate at which the water level [h] is rising when [h=2\ \text{m}]. (Volume [V=\pi r^{2}h].)

Step-by-step Solution:

1. Substitute radius: [V][=\pi(4)^{2}h][=16\pi h].
2. Differentiate w.r.t time: [\dfrac{dV}{dt}][=16\pi \dfrac{dh}{dt}].
3. Solve for [\dfrac{dh}{dt}]: [\dfrac{dh}{dt}][=\dfrac{1}{16\pi}\dfrac{dV}{dt}].
4. Substitute [\dfrac{dV}{dt}=10]: [\dfrac{dh}{dt}][=\dfrac{10}{16\pi}][=\dfrac{5}{8\pi}\ \text{m/s}].
5. Numerical approx (optional): [\dfrac{5}{8\pi}\approx 0.199\ \text{m/s}].

Conclusion: [\dfrac{dh}{dt}][=\dfrac{5}{8\pi}\ \text{m/s}] at [h=2\ \text{m}].

Question 4

A ladder 10 m long leans against a wall. The foot moves away from the wall at [\dfrac{dx}{dt}=0.6\ \text{m/s}]. Find how fast the top descends when the foot is 8 m from the wall. (Relation: [x^{2}+y^{2}=10^{2}].)

Step-by-step Solution:

1. Differentiate: [2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0].
2. Solve for [\dfrac{dy}{dt}]: [\dfrac{dy}{dt}][=-\dfrac{x}{y}\dfrac{dx}{dt}].
3. When [x=8], find [y][=\sqrt{100-64}][=\sqrt{36}=6].
4. Substitute values: [\dfrac{dy}{dt}][=-\dfrac{8}{6}\cdot 0.6 ][= -\dfrac{4}{3}\cdot 0.6 ][= -0.8\ \text{m/s}].

Conclusion: Top descends at [0.8\ \text{m/s}] (negative sign indicates downward): [\dfrac{dy}{dt}=-0.8\ \text{m/s}].

Question 5

A spherical balloon is being inflated so that its radius increases at [\dfrac{dr}{dt}=0.02\ \text{m/s}]. Find the rate of increase of the surface area when [r=0.5\ \text{m}]. (Surface area [S=4\pi r^{2}].)

Step-by-step Solution:

1. Differentiate area: [\dfrac{dS}{dt}][=\dfrac{d}{dt}(4\pi r^{2}) ][= 8\pi r \dfrac{dr}{dt}].
2. Substitute values: [\dfrac{dS}{dt}][=8\pi(0.5)(0.02) ][= 8\pi\cdot 0.01 ][= 0.08\pi\ \text{m}^{2}/\text{s}].
3. Numeric approx: [0.08\pi\approx 0.2513\ \text{m}^{2}/\text{s}].

Conclusion: [\dfrac{dS}{dt}][=0.08\pi\ \text{m}^{2}/\text{s}] at [r=0.5\ \text{m}].

Question 6

A particle moves so its velocity is given by [v(t)=\dfrac{d x}{dt}=4\sin t] (m/s). (a) Find the acceleration [\dfrac{d^{2}x}{dt^{2}}]. (b) Find acceleration at [t=\pi/6\ \text{s}].

Step-by-step Solution:

1. Acceleration is derivative of velocity: [\dfrac{d^{2}x}{dt^{2}} ][= \dfrac{d}{dt}(4\sin t) = 4\cos t].
2. Evaluate at [t=\pi/6]: [4\cos(\pi/6)][=4\cdot \dfrac{\sqrt{3}}{2}][=2\sqrt{3}\ \text{m/s}^{2}].

Conclusion: [\dfrac{d^{2}x}{dt^{2}}][=4\cos t,][\quad][ \dfrac{d^{2}x}{dt^{2}}\big|_{t=\pi/6}][=2\sqrt{3}\ \text{m/s}^{2}.]

Question 7

A conical tank (vertex down) has fixed height 9 m and base radius 3 m. Water is being poured so that the depth increases at [\dfrac{dh}{dt}=0.05\ \text{m/s}] when [h=6\ \text{m}]. Find the rate of change of the water volume at that moment. (Volume [V=\dfrac{1}{3}\pi r^{2}h]; similar triangles: [r/h = 3/9 = 1/3] so [r=\dfrac{h}{3}].)

Step-by-step Solution:

1. Express V in terms of h: [V][=\dfrac{1}{3}\pi\left(\dfrac{h}{3}\right)^{2}h ][= \dfrac{1}{3}\pi \dfrac{h^{2}}{9}h ][= \dfrac{1}{27}\pi h^{3}].
2. Differentiate: [\dfrac{dV}{dt} ][= \dfrac{1}{27}\pi\cdot 3h^{2}\dfrac{dh}{dt} ][= \dfrac{1}{9}\pi h^{2}\dfrac{dh}{dt}].
3. Substitute [h=6], [\dfrac{dh}{dt}=0.05]: [\dfrac{dV}{dt}][=\dfrac{1}{9}\pi(36)(0.05) ][= \dfrac{36}{9}\pi\cdot 0.05 ][= 4\pi\cdot 0.05 ][= 0.2\pi\ \text{m}^{3}/\text{s}].
4. Numeric approx: [0.2\pi][\approx 0.6283\ \text{m}^{3}/\text{s}].

Conclusion: [\dfrac{dV}{dt}][=0.2\pi\ \text{m}^{3}/\text{s}] at [h=6\ \text{m}].

Question 8

Light source at the ground casts a shadow. A 2 m tall man walks away from a light 6 m high at speed [\dfrac{dx}{dt}=1.2\ \text{m/s}]. Let [x] be distance of man from lamp, [y] be length of shadow. Find [\dfrac{dy}{dt}] when [x=4\ \text{m}]. (Use similarity: [\dfrac{6}{x+y} = \dfrac{2}{y}].)

Step-by-step Solution:

1. From similarity: [6y ][= 2(x+y) ][\Rightarrow 6y = 2x + 2y].
2. Rearr. : [6y – 2y = 2x ][\Rightarrow 4y = 2x ][\Rightarrow y = \dfrac{x}{2}].
3. Differentiate: [\dfrac{dy}{dt} ][= \dfrac{1}{2}\dfrac{dx}{dt}].
4. Substitute [\dfrac{dx}{dt}=1.2]: [\dfrac{dy}{dt} ][= \dfrac{1}{2}\cdot 1.2 ][= 0.6\ \text{m/s}].

Conclusion: Shadow lengthens at [0.6\ \text{m/s}] when [x=4\ \text{m}].

Question 9

An object moves in a circle of radius 5 m with angular speed [\omega(t)=\dfrac{d\theta}{dt}=0.2t] (rad/s). Find the rate of change of the arc length [s= r\theta] at time [t=3\ \text{s}].

Step-by-step Solution:

1. Arc length: [s=r\theta] so [\dfrac{ds}{dt}][=r\dfrac{d\theta}{dt}].
2. Substitute [r=5] and [\dfrac{d\theta}{dt}=0.2t]: [\dfrac{ds}{dt}=5\cdot 0.2t = t].
3. At [t=3]: [\dfrac{ds}{dt}=3\ \text{m/s}].

Conclusion: Arc length increases at [3\ \text{m/s}] at [t=3\ \text{s}].

Question 10

A hemispherical tank (radius 2 m) is being filled so that the water depth measured from bottom is increasing at [\dfrac{dh}{dt}=0.025\ \text{m/s}] when [h=1\ \text{m}]. Express volume of water as function of h and find [\dfrac{dV}{dt}] at that instant. (Volume of spherical cap of height h on a sphere of radius R is [V=\pi h^{2}\big(R-\dfrac{h}{3}\big)]. For hemisphere where water depth h ≤ R use same cap formula with [R=2].)

Step-by-step Solution:

1. Use cap formula with [R=2]: [V(h)][=\pi h^{2}\Big(2 – \dfrac{h}{3}\Big)].
2. Expand or differentiate using product rule:
   [\dfrac{dV}{dt} ][= \dfrac{dV}{dh}\dfrac{dh}{dt}].
3. Compute [\dfrac{dV}{dh}]:
   [\dfrac{dV}{dh} ][= \pi\Big(2\cdot h\big(2-\dfrac{h}{3}\big) + h^{2}\big(-\dfrac{1}{3}\big)\Big)] (product rule simplified).
   Simplify inside: first term = [2h(2 – h/3)][=4h – \dfrac{2h^{2}}{3}]; add second term = [ -\dfrac{h^{2}}{3} ] total = [4h – \dfrac{2h^{2}}{3} – \dfrac{h^{2}}{3}][ = 4h – h^{2}].
   So [\dfrac{dV}{dh} ][= \pi(4h – h^{2})].
4. Now [\dfrac{dV}{dt} ][= \pi(4h – h^{2})\dfrac{dh}{dt}].
5. Substitute [h=1], [\dfrac{dh}{dt}=0.025]: [\dfrac{dV}{dt} ][= \pi(4 – 1)\cdot 0.025 ][= \pi\cdot 3 \cdot 0.025 ][= 0.075\pi\ \text{m}^{3}/\text{s}].
6. Numeric approx: [0.075\pi][\approx 0.2356\ \text{m}^{3}/\text{s}].

Conclusion: [\dfrac{dV}{dt}][=0.075\pi\ \text{m}^{3}/\text{s}] when [h=1\ \text{m}].