A Collection of Problems in Differential Calculus: Problems from Calculus I Final Examinations, 2000-2020 Department of Mathematics, Simon Fraser University

Section 6.5 Related Rates

These answers correspond to the problems in Section 2.3.

1. Solve for \(x\) in the system of equations: \(\ds \frac{dy}{dt}=2x\frac{dx}{dt}, \ \frac{dy}{dt}=\frac{dx}{dt}\text{.}\) \(\ds x=\frac{1}{2},\ y=\frac{1}{4}\text{.}\)

2. Let \(\alpha\) be the angle between the \(x\)-axis and the line \(l\text{.}\) Then \(\ds \tan \alpha = x-4+\frac{8}{x}\) and \(\ds\sec^2\alpha\cdot\frac{d\alpha}{dt}=\left(1-\frac{8}{x^2}\right)\cdot \frac{dx}{dt}\text{.}\) Therefore, \(\ds \left. \frac{d\alpha}{dt}\right|_{x=3}=-\frac{3}{17}\) rad/s.

3. Let \(x = x(t)\) be the distance between the wall and the child. Let \(s = s(t)\) be the length of the child's shadow. Using similar triangles, it is found that \(\ds \frac{10}{1.5}=\frac{x+s}{s}\text{.}\) It follows that \(\ds \frac{ds}{dt}=\frac{3}{17}\cdot \frac{dx}{dt}\text{.}\) \(\ds \left.\frac{ds}{dt}\right|_{x=4}= -0.44\)m/s.

4. Let \(x = x(t)\) be the distance between the light and the wall in metres. Let \(p = p(t)\) be the height of the man's shadow in metres. Using similar triangles it is found that \(\ds \frac{p-1}{1}=\frac{x}{x-4}\text{.}\) It follows that \(\ds \frac{dp}{dt}=-\frac{4}{(x-4)^2} \cdot \frac{dx}{dt}\text{.}\) \(\ds \left. \frac{dp}{dt}\right|_{x=8}=0.5\)m/s.

5. Let \(x=x(t)\) be the distance between the bottom of the ladder and the wall. It is given that, at any time \(t\text{,}\) \(\ds \frac{dx}{dt}=2\) ft/s. Let \(\theta =\theta (t)\) be the angle between the top of the ladder and the wall. Then \(\ds \sin \theta =\frac{x(t)}{15}\text{.}\) It follows that \(\ds \cos \theta \cdot \frac{d\theta }{dt}=\frac{1}{15}\frac{dx}{dt}\text{.}\) Thus when \(\ds \theta = \frac{\pi }{3}\) the rate of change of \(\theta\) is given by \(\ds \frac{d\theta }{dt}=\frac{4}{15}\) ft/s.

6. Let \(x=x(t)\) be the distance between the foot of the ladder and the wall and let \(y=y(t)\) be the distance between the top of the ladder and the ground. It is given that, at any time \(t\text{,}\) \(\ds \frac{dx}{dt}=\frac{1}{2}\) m/min. From \(\ds x^2+y^2 =144\) it follows that \(\ds x \cdot \frac{dx}{dt}+ y\frac{dy}{dt}=0\text{.}\) Thus when \(\ds x(t) =4\) we have that \(y(t)=8\sqrt{2}\) and \(\ds 4\cdot \frac{1}{2}+8\sqrt{2}\frac{dy}{dt}=0\text{.}\) The top of the ladder is falling at the rate \(\ds \frac{dy}{dt}=-\frac{\sqrt{2}}{8}\) m/min.

7. Let \(x=x(t)\) be the height of the rocket at time \(t\) and let \(y=y(t)\) be the distance between the rocket and radar station. It is given that, at any time \(t\text{,}\) \(x^2= y^2-16\text{.}\) Thus, at any time \(t\text{,}\) \(\ds x \cdot \frac{dx}{dt}= y\frac{dy}{dt}\text{.}\) At the instant when \(y=5\) miles and \(\ds \frac{dy}{dt}=3600\) mi/h we have that \(x=3\) miles and we conclude that, at that instant, \(\ds 3\frac{dx}{dt}=5\cdot 3600\text{.}\) Thus the vertical speed of the rocket is \(\ds v= \frac{dx}{dt}=6000\) mi/h.

8. Let \(x=x(t)\) be the distance between the dock and the bow of the boat at time \(t\) and let \(y=y(t)\) be the length of the rope between the pulley and the bow at time \(t\text{.}\) It is given that \(\ds \frac{dy}{dt}=1\) m/sec. From \(x^2+1=y^2\) it follows that \(\ds \frac{dx}{dt}=\frac{y}{x}\) m/sec. Since \(y=10\) implies \(x=\sqrt{99}\) we conclude that when 10 m of rope is out then the boat is approaching the dock at the rate of \(\ds \frac{10}{\sqrt{99}}\) m/sec.

9. (a) Let the positive direction of the \(x\)-axis be the line of the movement of the ship. If \((x(t),0)\) is the position of the ship at time \(t\) then the position of the submarine is given by \((x(t),y(t))\) with \(\ds y(t)=-\frac{\sqrt{3}}{3}\cdot x(t)+c\text{,}\) for some negative constant \(c\text{.}\) See Figure 6.5.1. (b) \(\ds \frac{dy}{dt}=-\frac{10\sqrt{3}}{3}\) km/h.

   

10. From \(\ds y=5\tan \theta\) we get that, at any time \(t\text{,}\) \(\ds \frac{dy}{dt}=5\sec ^2\theta \frac{d\theta }{dt}\text{.}\) At the instant when \(\ds \theta =\frac{\pi }{3}\) radians we have that \(\ds v=\frac{dy}{dt}=5\cdot \sec ^2 \frac{\pi }{3} \cdot 2=40\) m/s.

11. Let \(x = x(t)\) be the horizontal distance (in km) of the train with respect to the camera. Let \(D = D(t)\) be the distance (in km) between the camera and the train. From the relationship \(\ds \tan \theta=\frac{x}{0.5}\) it follows that \(\ds \frac{d\theta }{dt}=2\cos^2\theta \cdot\frac{dx}{dt}\text{.}\) When \(D=1\text{,}\) \(\cos \theta=2\text{,}\) and therefore \(\ds \left.\frac{d\theta}{dt}\right|_{D=1}=1.5\) rad/min.

12. After time \(t\) (in hours) the plane is \(480t\) km away from the point directly above the observer. Thus, at time \(t\text{,}\) the distance between the observer and the plane is \(\ds D=\sqrt{3^2+(480t)^2}\text{.}\) We differentiate \(D^2=9+230,400t^2\) with respect to \(t\) to get \(\ds 2D\frac{dD}{dt}=460,800t\text{.}\) Since \(\ds 30\mbox{ sec} =\frac{1}{120}\) hours it follows that the distance between the observer and the plane after 30 seconds equals \(D=5\) km. Thus, 30 seconds later the distance \(D\) from the observer to the airplane is increasing at the rate of \(\ds \left. \frac{dD}{dt}\right| _{t=\frac{1}{120}}= 384\) km/h.

13. Let \(y\) be the distance between the airplane and the radar station. Then, as the hypothenuse in a right angle triangle with the angle \(\theta\) and the opposite leg of length 1000 m, \(\ds y=\frac{1000}{\sin \theta }\text{.}\) Since it is given that \(\ds \frac{d\theta }{dt}=-0.1\) rad/sec, it follows that \(\ds \frac{dy}{dt}=-\frac{1000\cos \theta }{\sin ^2\theta }\cdot \frac{d\theta }{dt}=\frac{100\cos \theta }{\sin ^2\theta }\) m/sec. Hence if \(\ds \theta =\frac{\pi }{4}\text{,}\) the speed of the plane is given by \(\ds \left. \frac{dy}{dt}\right| _{t=\frac{\pi }{4}}=100\sqrt{2}\) m/sec.

14. (a) From \(z^2=64+4t^2\) it follows that \(2zz'=8t\text{.}\) If \(z=10\) then \(t=3\) and at that instant \(z'=1.2\) m/s. (b) Since the height of the kite after \(t\) seconds is \(2t\) metres, it follows that \(\ds \tan x=\frac{2t}{8}\text{.}\) Thus \(\ds \frac{x'}{\cos ^2x}=\frac{1}{4}\text{.}\) If \(y=6\) then \(t=3\) and \(\ds \tan x=\frac{3}{4}\text{.}\) It follows that \(\ds \cos x=\frac{4}{5}\) and at that instant the rate of change of \(x\) is given by \(\ds x'=x'(3)=\frac{4}{25}\) m/s.

15. Let \(D = D(t)\) be the distance between the girl's hand and the kite in feet. From the relationship \(D^2=x^2+ y^2\text{,}\) it follows that \(\ds D\frac{dD}{dt}=x \frac{dx}{dt}+ y \frac{dy}{dt}\text{.}\) When \(D=125\)ft, \(y=100\)ft and \(x=75\)ft, and \(\ds \left.\frac{dx}{dt}\right|_{D=125}=3.33\) ft/s.

16. Let \(x=x(t)\) be the distance (in metres) between the boy and the balloon at time \(t\text{.}\) Then \([x(t)]^2=(8t)^2+(36+4t)^2\text{.}\) From \(2x(t)x^\prime (t)=128t+8(36+4t)\text{.}\) From \(x(3)=24\sqrt{5}\) m, it follows that \(\ds x^\prime (3)=\frac{16}{\sqrt{5}}\) m/sec.

17. Let \(D = D(t)\) be the distance (in metres) between the girl and the balloon at time \(t\text{.}\) Let \(x=x(t)\) and \(y=y(t)\) be the horizontal and vertical distances to the balloon respectively. Observe that, for \(t>0\text{,}\) \(\frac{dx}{dt}=-4\) m/sec and \(\frac{dy}{dt}=2\) m/sec. From the relationship \(D^2=x^2+ y^2\text{,}\) it follows that \(\ds D \frac{dD}{dt}=x \frac{dx}{dt}+ y \frac{dy}{dt}\text{.}\) At time \(t = 2\text{,}\) \(x = 2\text{,}\) \(y =y(0)+ 4\text{,}\) and \(D= \sqrt{4+(y(0)+4)^2}\text{.}\) Therefore, \(\ds \left.\frac{dD}{dt}\right|_{t=2}=\frac{2y(0)}{\sqrt{4+(y(0)+4)^2}}\)m/sec.

18. Let \(\theta = \theta (t)\) be the elevation angle. From \(\ds \tan \theta =\frac{2t}{80}\) it follows that \(\ds \frac{d\theta }{dt}=\frac{\cos ^2\theta }{40}\text{.}\) When \(t=30\) we have \(\ds \tan \theta =\frac{3}{4}\) and \(\ds \cos \theta =\frac{4}{5}\text{.}\) Thus when the helicopter is 60 m above the ground the elevation angle of the observer's line of sight to the helicopter is changing at the rate \(\ds \frac{1}{50}\) m/sec.

19. Let \(r\) denotes the radius of the circular containment area. It is given that \(\ds \frac{dr}{dt}=-5\) m/min. From the fact that the area at time \(t\) is given by \(A=r^2\pi\text{,}\) where \(r=r(t)\text{,}\) it follows that \(\ds \frac{dA}{dt} =2r\pi\frac{dr}{dt}=-10r\pi\) m\(^2\)/min. Hence when \(r=50\)m then the area shrinks at the rate of \(10\cdot50\cdot \pi=500\pi\) m\(^2\)/min.

20. The area \(A = A(t)\) of the inscribed rectangle can be described as \(A=4xy\text{,}\) where \(x\) and \(y\) are the coordinates of point \(P\text{.}\) Recall that \(x=\cos \theta\) and \(y=\sin \theta\text{.}\) It follows that \(A(t)=2\sin (2\theta)\) and \(\ds \frac{dA}{dt}=4\cos(2\theta)\cdot\frac{d\theta}{dt}\text{.}\) Therefore, \(\ds \left.\frac{dA}{dt}\right|_{\theta=\frac{\pi}{3}}=-4\) cm\(^2\)/sec.

21. Let \(x=x(t)\) be the edge length. Then the volume is given by \(V=x^3\) and the surface area is given by \(S=6x^2\text{.}\) It is given that \(\ds \frac{dV}{dt}=10\text{.}\) This implies that \(\ds 3x^2\frac{dx}{dt}=10\) at any time \(t\) and we conclude that at the instant when \(x=8\) the edge is increasing at the rate \(\ds \frac{5}{96}\)cm/min. This fact together with \(\ds \frac{dS}{dt}=12x\frac{dx}{dt}\) implies that at the instant when \(x=8\) the surface area is increasing at the rate \(\ds 12\cdot 8\cdot \frac{5}{96}=5\) cm\(^2\)/min.

22. Let \(x = x(t)\) be the edge length of the cube in cm. Given that the volume, \(V=x^3\) it follows that \(\ds \frac{dV}{dt}=3x^2 \frac{dx}{dt}\text{.}\) The surface area \(S = S(t)\) is described by \(S=6x^2\) leading to the relationship \(\ds \frac{dx}{dt}= \frac{1}{12x} \cdot \frac{dS}{dt}\text{.}\) Therefore, \(\ds \frac{dV}{dt}=\frac{x}{4}\cdot \frac{dS}{dt}\) and \(\ds \left.\frac{dV}{dt}\right|_{x=4}=7.5\) cm\(^3\)/min.

23. Let \(x = x(t)\) be the edge length of the cube in cm. The length of the diagonal, \(z = z(x)\) can be described as \(z=\sqrt{3} x\text{.}\) It follows that \(\ds \frac{dz}{dt}=\sqrt{3}\frac{dx}{dt}=-3\sqrt{3}\) cm/sec.

24. Let \(x=x(t)\) be the side length of the ice cube. Since the volume of the cube can be expressed as \(V = x^3\text{,}\) it follows that \(\ds \frac{dx}{dt}=\frac{1}{3x^2}\frac{dV}{dt}\text{.}\) \(\ds \left.\frac{dx}{dt}\right|_{x=2}=-2.5\)m/sec

25. Let \(H=H(t)\) be the height of the box, let \(x=x(t)\) be the length of a side of the base, and \(V=V(t)=Hx^2\text{.}\) It is given that \(\ds \frac{dH}{dt}=2\) m/sec and \(\ds \frac{dV}{dt}=2x\frac{dx}{dt}H+x^2\frac{dH}{dt}=-5\) m\(^3\)/sec. The question is to find the value of \(\ds \frac{dx}{dt}\) at the instant when \(x^2=64\) m\(^2\) and \(H=8\) m. Thus, at that instant, one of the sides of the base is decreasing at the rate of \(\ds \frac{133}{128}\) m/sec.

26. (a) The volume of the water is expressed as \(\ds V=\frac{1}{3} \pi r^2 h\text{.}\) The radius \(r\) can be represented as \(\ds r=\frac{Rh}{H}\) using similar triangles. Overall, \(\ds V=\frac{\pi}{3} \frac{R^2h^3}{H^2}\text{.}\) (b) From \(\ds \frac{dV}{dt}=\pi h^2\left(\frac{R}{H}\right)^2\frac{dh}{dt}\) it follows \(\ds \frac{R}{H}=\frac{1}{8}\sqrt{\frac{5}{\pi}}\text{.}\)

27. Let \(H=H(t)\) be the height of the pile, let \(r=r(t)\) be the radius of the base, and let \(V=V(t)\) be the volume of the cone. It is given that \(H=r\) (which implies that \(\ds V=\frac{H^3\pi }{3}\)) and that \(\ds \frac{dV}{dt}=H^2\pi \frac{dH}{dt}=1\) m\(^3\)/sec. The question is to find the value of \(\ds \frac{dH}{dt}\) at the instant when \(H=2\text{.}\) Thus at that instant the sandpile is rising at the rate of \(\ds \frac{1}{4\pi }\) m/sec.

28. Let \(x = x(t)\) be the height and diameter of the gravel pile. The volume of this pile can be expressed as \(\ds V=\frac{\pi x^2}{12}\text{.}\) Given \(\ds \left.\frac{dV}{dt}\right|_{x=5}=1\) m\(^3\)/sec it follows that \(\ds \left. \frac{dx}{dt}\right|_{x=5}=\frac{4}{25\pi}\) m/sec.

29. Let \(H=H(t)\) be the height of water, let \(r=r(t)\) be the radius of the surface of water, and let \(V=V(t)\) be the volume of water in the cone at time \(t\text{.}\) It is given that \(\ds r=\frac{3H}{5}\) which implies that \(\ds V=\frac{3H^3\pi }{25}\text{.}\) The question is to find the value of \(\ds \frac{dH}{dt}\) at the instant when \(H=3\) and \(\ds \frac{dV}{dt}=-2\) m\(^3\)/sec . Thus at that instant the water level dropping at the rate of \(\ds \frac{50}{81\pi }\) m/sec.

30. Let \(V=V(t)\) be the volume of water in the cone at time \(t\text{,}\) \(h=h(t)\) be the height of water in the cone, \(w=w(t)\) be the volume of water in the cylinder and \(H=H(t)\) be the height of water in the cylinder. The volume of water in the cone can be expressed as \(\ds V=\frac{16}{75} \pi h^3\) leading to \(\ds \frac{dV}{dt}=-\frac{dw}{dt}=\frac{48}{75} \pi h^2 \frac{dh}{dt}= -\frac{72}{25 \pi}\) m\(^3\)/min. The volume of water in the cylinder can be expressed as \(\ds w=\pi r^2 h\) leading to \(\ds \frac{dh}{dt}=\frac{1}{16\pi} \frac{dw}{dt}=\frac{9}{50}\) m/min.

31. The distance between the boy and the girl is given by \(z=\sqrt{x^2+y^2}\) where \(x=x(t)\) and \(y=y(t)\) are the distances covered by the boy and the girl in time \(t\text{,}\) respectively. The question is to find \(z'(6)\text{.}\) We differentiate \(z^2=x^2+y^2\) to get \(zz'=xx'-yy'\text{.}\) From \(x(6)=9\text{,}\) \(y(6)=12\text{,}\) \(z(6)=15\text{,}\) \(x'(t)=1.5\text{,}\) and \(y'(t)=2\) it follows that \(\ds z'(6)=2.5\) m/s.

32. The distance between the two ships is given by \(z=\sqrt{x^2+(60-y)^2}\) where \(x=x(t)\) and \(y=y(t)\) are the distances covered by the ship \(A\) and the ship \(B\) in time \(t\text{,}\) respectively. The question is to find \(z'(4)\text{.}\) We differentiate \(z^2=x^2+(60-y)^2\) to get \(zz'=xx'-(60-y)y'\text{.}\) From \(x(4)=60\text{,}\) \(y(4)=49\text{,}\) \(z(4)=61\text{,}\) \(x'(t)=15\text{,}\) and \(y'(t)=12.25\) it follows that \(\ds z'(4)=\frac{765.25}{61}\approx 12.54\) miles/hour.

33. Let the point \(L\) represents the lighthouse, let at time \(t\) the light beam shines on the point \(A=A(t)\) on the shoreline, and let \(x=x(t)\) be the distance between \(A\) and \(P\text{.}\) Let \(\theta = \theta (t)\) be the measure in radians of \(\angle{PLA}\text{.}\) It is given that \(x=3\tan \theta\) and \(\ds \frac{d\theta }{dt} =8\pi\) radians/minute. The question is to find \(\ds \frac{dx}{dt}\) at the instant when \(x=1\text{.}\) First we note that \(\ds \frac{dx}{dt}=3\sec ^2\theta \frac{d\theta }{dt}\text{.}\) Secondly, at the instant when \(x=1\) we have that \(\ds \tan \theta =\frac{1}{3}\) which implies that \(\ds \cos \theta =\frac{3}{\sqrt{10}}\text{.}\) Hence, when shining on a point one kilometre away from \(P\text{,}\) the light beam moving along the shoreline at the rate of \(\ds \frac{80\pi }{3}\) km/min.

34. Let \(z=z(t)\) be the distance between the two puppies at time \(t\text{.}\) From \(z^2=x^2+y^2\) it follows that \(\ds z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\) and \(\ds \left.\frac{dz}{dt}\right|_{z=5}=\frac{17}{5}\) m/sec.

35. Let \(z = z(t)\) be the distance between the runner and cyclist in metres at time \(t\text{.}\) Let \(x = x(t)\) be the horizontal distance between the two individuals. Then \(z(t)\) can be expressed as \(z^2=x^2+ 900\) which leads to \(\ds z\frac{dz}{dt}=x \frac{dx}{dt}\text{.}\) The reduction in horizontal distance between the two individuals is given by\(\ds \frac{dx}{dt}=-5-2=-7\) m/sec. Observe that when \(t=1\) minute then \(x=80\) m and \(\ds z=10\sqrt{73}\) m. \(\ds \left.\frac{dz}{dt}\right|_{t=1 \text{ min } }=-\frac{56}{\sqrt{73}}\) m/sec.

36. Let \(x=x(t)\) be the distance between the police car and the intersection and let \(y=y(t)\) be the distance between the SUV and the intersection. The distance between the two cars is given by \(z=\sqrt{x^2+y^2}\text{.}\) The question is to find the value of \(\ds \frac{dy}{dt}\) at the instant when \(x=0.6\) km, \(y=0.8\) km, \(\ds \frac{dz}{dt}=20\) km/hr, and \(\ds \frac{dx}{dt}=-60\) km/hr. We differentiate \(z^2=x^2+y^2\) to get \(\ds z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\text{.}\) Since, at the given instance, \(z=1\text{,}\) we have that \(\ds \frac{dy}{dt}= 70\) km/hr.

37. Let \(x = x(t)\) be the horizontal and let \(y = y(t)\) be the vertical distance of the rider from the centre of the Ferris wheel (in metres). From the relationship \(y=10\cos \theta\) it follows that \(\ds \frac{dy}{dt}=-10\sin \theta \cdot \frac{d\theta}{dt}\text{.}\) When \(x=6\) m then \(\ds \sin \theta =\frac{3}{5}\text{.}\) Therefore, \(\ds \left.\frac{dy}{dt}\right|_{x=6}=12\) m/sec.

38. (a) Observe that \(\ds \frac{d\theta}{dt}=12\pi\) rad/sec. Using the Law of Sines we find that \(3\sin \alpha =\sin \theta\text{.}\) Hence \(\ds 3\cos \alpha \cdot \frac{d\alpha}{dt}= \cos \theta \cdot \frac{d\theta}{dt}\text{.}\) If \(\ds \theta =\frac{\pi}{3}\) then \(\ds \sin\alpha =\frac{\sqrt{3}}{6}\) and \(\ds \left.\frac{d\alpha}{dt}\right|_{\theta=\frac{\pi}{3}}=\frac{12\pi}{\sqrt{33}}\) rad/sec. (b) By the Law of Cosines \(120^2=x^2+40^2-80x\cos\theta\text{.}\) It follows that \(x=40(\cos \theta+\sqrt{8+\cos \theta})\text{.}\) (c) \(\ds \frac{dx}{dt}=-40\left(1+\frac{\cos \theta}{\sqrt{8+\cos^2\theta}}\right)\cdot \sin\theta\) rad/sec.