Exercises 2.3 Related Rates
To solve a related rates problem you need to do the following:
Identify the independent variable on which the other quantities depend and assign it a symbol, such as \(t\text{.}\) Also, assign symbols to the variable quantities that depend on \(t\text{.}\)
Find an equation that relates the dependent variables.
Differentiate both sides of the equation with respect to \(t\) (using the chain rule if necessary).
Substitute the given information into the related rates equation and solve for the unknown rate.
Solve the following related rates problems:
1.
A bug is walking on the parabola \(y=x^2\text{.}\) At what point on the parabola are the \(x-\) and \(y-\) coordinates changing at the same rate?
2.
A particle is moving along the parabola \(y=x^2-4x+8\text{.}\) Its \(x-\)coordinate as a function of time \(t\) (in seconds) is \(x(t)=-2t^3+5\) metres. Let \(l\) be the line joining the origin \((0,0)\) to the particle. Determine how quickly the angle between the \(x\)-axis and the line \(l\) is changing when \(x=3\text{.}\)
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.
A child 1.5 m tall walks towards a lamppost on the level ground at the rate \(0.25\) m/sec. The lamppost is 10 m high. How fast is the length of the child's shadow decreasing when the child is 4 m from the post?
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 \(\ds \frac{ds}{dt}=\frac{3}{17}\cdot \frac{dx}{dt}\) and \(\ds \left.\frac{ds}{dt}\right|_{x=4}= -0.044\)m/s.
4.
A light moving at \(2\) m/sec approaches a 2-m tall man standing 4 m from a wall. The light is 1 m above the ground. How fast is the tip \(P\) of the man's shadow moving up the wall when the light is 8 m from the wall? )
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.
A ladder 15 ft long rests against a vertical wall. Its top slides down the wall while its bottom moves away along the level ground at a speed of 2 ft/s. How fast is the angle between the top of the ladder and the wall changing when the angle is \(\pi /3\) radians?
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.
A ladder 12 metres long leans against a wall. The foot of the ladder is pulled away from the wall at the rate \(\frac{1}{2}\) m/min. At what rate is the top of the ladder falling when the foot of the ladder is 4 metres from the wall?
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.
A rocket \(R\) is launched vertically and it is tracked from a radar station \(S\) which is \(4\) miles away from the launch site at the same height above sea level. At a certain instant after launch, \(R\) is 5 miles away from \(S\) and the distance from \(R\) to \(S\) is increasing at a rate of 3600 miles per hour. Compute the vertical speed \(v\) of the rocket at this instant.
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.
A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the bow of the boat at a point 1 m below the pulley. If the rope is pulled through the pulley at a rate of 1 m/sec, at what rate will the boat be approaching the dock when 10 m of rope is out?
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 ship is moving on the surface of the ocean in a straight line at 10 km/hr. At the same time, an enemy submarine maintains a position directly below the ship while diving at an angle of 30 degrees below the horizontal.
provide a diagram of this problem situation. Use \(x\) as the line of movement of the ship in kilometres and \(y\) as the depth of the submarine in kilometres.
How fast is the submarine depth increasing?
let \(c \le 0\text{.}\)
\(-\frac{10\sqrt{3}}{3}\) km/hr.
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 \(c\text{.}\)
We have that 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 \(c\text{.}\) Therefore, \(\displaystyle \frac{dy}{dt} = -\frac{\sqrt{3}}{3} \frac{dx}{dt} = =-\frac{10\sqrt{3}}{3}\) km/hr.
10.
A person (\(A\)) situated at the edge of the river observes the passage of a speed boat going downstream. The boat travels exactly through the middle of the river (at the distance \(d\) from the riverbank.) The river is 10 m wide. When the boat is at \(\theta =60^0\) the observer measures the rate of change of the angle \(\theta\) to be 2 radians/second. What is the speed, \(v\text{,}\) of the speed boat at that instant?
11.
A high speed train is traveling at 3 km/min along a straight track. The train is moving away from a movie camera which is located 0.5 km from the track. The camera keeps turning so as to always point at the front of the train. How fast (in radians per minute) is the camera rotating when the train is 1 km from the camera?
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.
An airplane is flying horizontally at an altitude of \(y=3\) km and at a speed of 480 km/h passes directly above an observer on the ground. How fast is the distance \(D\) from the observer to the airplane increasing 30 seconds later?
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.
An airplane flying horizontally at a constant height of \(1000\) m above a fixed radar station. At a certain instant the angle of elevation \(\theta\) from the station is \(\ds \frac{\pi }{4}\) radians and decreasing at a rate of 0.1 rad/sec. What is the speed of the aircraft at this moment.
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 kite is rising vertically at a constant speed of 2 m/s from a location at ground level which is 8 m away from the person handling the string of the kite.
Let \(z\) be the distance from the kite to the person. Find the rate of change of \(z\) with respect to time \(t\) when \(z=10\text{.}\)
Let \(x\) be the angle the string makes with the horizontal. Find the rate of change of \(x\) with respect to time \(t\) when the kite is \(y=6\) m above ground.
\(1.2\) m/s.
\(\frac{4}{25}\) m/s.
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.
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.
A girl flying a kite holds the string 5 feet above the ground level and lets the string out at a rate of 2 ft/sec as the kite moves horizontally at an altitude of 105 feet. Assuming there is no sag in the string, find the rate at which the kite is moving when 125 feet of string has been let out.
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.
A balloon is rising at a constant speed 4m/sec. A boy is cycling along a straight road at a speed of 8m/sec. When he passes under the balloon, it is 36 metres above him. How fast is the distance between the boy and balloon increasing 3 seconds later.
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.
A helicopter takes off from a point 80 m away from an observer located on the ground, and rises vertically at 2 m/s. At what rate is the elevation angle of the observer's line of sight to the helicopter changing when the helicopter is 60 m above the ground.
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.
18.
An oil slick on a lake is surrounded by a floating circular containment boom. As the boom is pulled in, the circular containment area shrinks (all the while maintaining the shape of a circle.) If the boom is pulled in at the rate of 5 m/min, at what rate is the containment area shrinking when it has a diameter of 100m?
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.
19.
A rectangle is inscribed in the unit circle so that its sides are parallel to the coordinate axis. Let \(\theta \in \left(0,\frac{\pi}{2}\right)\) be the angle between the positive \(x\)-axis and the ray with the initial point at the origin and passing through the top-right vertex \(P\) of the rectangle. Suppose that the angle \(\theta\) is increasing at the rate of \(2\mbox{ radians} /\mbox{second}\) and suppose that all lengths are measured in centimetres. At which rate is the area of the rectangle changing when \(\displaystyle \theta =\frac{\pi}{3}\text{?}\) Is the area increasing or decreasing at that moment? Why? Show all your work. Do not forget to use the appropriate units. Clearly explain your reasoning.
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.
20.
Consider a cube of variable size. (The edge length is increasing.) Assume that the volume of the cube is increasing at the rate of 10 cm\(^3\)/minute. How fast is the surface area increasing when the edge length is 8 cm?
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.
21.
Consider a cube of variable size. (The edge length is increasing.) Assume that the surface area of the cube is increasing at the rate of \(6\) cm\(^2\)/minute. How fast is the volume increasing when the edge length is 5 cm?
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.
22.
At what rate is the diagonal of a cube changing if its edges are decreasing at a rate of \(3\) cm/s?
23.
The volume of an ice cube is decreasing at a rate of 5 m\(^3\)/s. What is the rate of change of the side length at the instant when the side lengths are 2 m?
24.
The height of a rectangular box is increasing at a rate of 2 metres per second while the volume is decreasing at a rate of 5 cubic metres per second. If the base of the box is a square, at what rate is one of the sides of the base decreasing, at the moment when the base area is 64 square metres and the height is 8 metres?
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.
25.
A coffee filter has the shape of an inverted cone with a fixed top radius \(R\) and height \(H\text{.}\) Water drains out of the filter at a rate of 10 cm\(^3\)/min. When the depth \(h\) of the water is 8 cm, the depth is decreasing at a rate of 2 cm/min.
Express the volume of the cone as a function of the depth of the water only.
What is the ratio \(R/H\text{?}\)
\(\ds V=\frac{\pi}{3} \frac{R^2h^3}{H^2}\text{.}\)
\(\ds \frac{1}{8}\sqrt{\frac{5}{\pi}}\text{.}\)
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, \(\ds V=\frac{\pi}{3} \frac{R^2h^3}{H^2}\text{.}\)
From \(\ds \frac{dV}{dt}=\pi h^2\left(\frac{R}{H}\right)^2\frac{dh}{dt}\) it follows \(\ds \frac{R}{H}=\frac{1}{8}\sqrt{\frac{5}{\pi}}\text{.}\)
26.
Sand is pouring out of a tube at 1 cubic metre per second. It forms a pile which has the shape of a cone. The height of the cone is equal to the radius of the circle at its base. How fast is the sandpile rising when it is 2 metres high?
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.
27.
Gravel is being dumped from a conveyer belt at a rate of 1 cubic metre per second. It makes a pile in the the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile changing when it is 5 metres high?
Let \(x = x(t)\) be the height and diameter of the gravel pile. The volume of this pile can be expressed as \(\ds V=\frac{\pi x^3}{12}\text{.}\) 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.
28.
A water tank is in the shape of a cone with its vertex pointing downwards. The tank has a radius of 3 m and is 5 m high. At first the tank is full of water, but at time \(t=0\) (in seconds), a small hole at the vertex is opened and the water begins to drain. When the height of water in the tank has dropped to 3 m, the water is flowing out at 2 m\(^3\)/s. At what rate, in metres per second, is the water level dropping then?
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.
29.
A conical tank with an upper radius of 4 m and a height of 5 m drains into a cylindrical tank with radius of 4 m and a height of 5 m. The water level in the conical tank is dropping at a rate of 0.5 m/min when the water level of the conical tank is 3m. At what rate is the water level in the cylindrical tank rising at that point?
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.
30.
A boy starts walking north at a speed of 1.5 m/s, and a girl starts walking west from the same point \(P\) at the same time at a speed of 2 m/s. At what rate is the distance between the boy and the girl increasing 6 seconds later?
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.
31.
At noon of a certain day, ship \(A\) is 60 miles due north of ship \(B\text{.}\) If ship \(A\) sails east at speed of 15 miles per hour and \(B\) sails north at speed of 12.25 miles per hour, determine how rapidly the distance between them is changing 4 hours later?
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.
32.
A police car, approaching a right-angled intersection from the north, is chasing a speeding SUV that has turned the corner and is now moving straight east. When the police car is 0.6 km north of the intersection and the SUV is 0.8 km east of the intersection, the police determine with radar that the distance between them and the SUV is increasing at 20 km/hr. If the police car is moving at 60 km/hr at the instant of measurement, what is the speed of the SUV?
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.
33.
A ball lands at a point \(A\text{.}\) As soon as the ball lands, Puppy \(\#1\) starts off 15m North of \(A\) running at 3 m/s in the direction of \(A\text{,}\) and Puppy \(\#2\) starts off 12m East of \(A\) running at 2 m/s in the direction of \(A\text{.}\) At what rate is the distance between the puppies changing when they are \(5\)m apart. In answering this question, sketch a diagram and define your terms clearly.
34.
A person is walking east along a river bank path at a speed of \(2m/s\text{.}\) A cyclist is on a path on the opposite bank cycling west at a speed of \(5 m/s\text{.}\) The cyclist is initially 500\(m\) east of the walker. If the paths are \(30m\) apart how fast is the distance between the walker and cyclist changing after one minute?
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.
35.
A lighthouse is located on a small island 3 km off-shore from the nearest point \(P\) on a straight shoreline. Its light makes 4 revolutions per minute. How fast is the light beam moving along the shoreline when it is shining on a point 1 km along the shoreline from \(P\text{?}\)
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.
36.
You are riding on a ferris wheel of diameter 20 metres. The wheel is rotating at 1 revolution per minute. How fast are you rising when you are at the point \(P\) in the Figure below, that is you are 6 metres horizontally away from the vertical line passing the centre of the wheel?
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.
37.
The Figure below shows a rotating wheel with radius \(40\) cm and a connecting rod \(AP\) with length \(1.2\) m. The pin \(P\) slides back and forth along the \(x\)-axis as the wheel rotates counterclockwise at a rate of \(360\) revolutions per minute.
Find the angular velocity of the connecting rod, \(\frac{d\alpha}{dt}\text{,}\) in radians per second, when \(\theta=\frac{\pi}{3}\text{.}\)
Express the distance \(x=|OP|\) in terms of \(\theta\text{.}\)
Find an expression for the velocity of the pin \(P\) in terms of \(\theta\text{.}\)
\(\frac{12\pi}{\sqrt{33}}\) rad/sec.
\(x=40(\cos \theta+\sqrt{8+\cos \theta})\text{.}\)
\(\ds -40\left(1+\frac{\cos \theta}{\sqrt{8+\cos^2\theta}}\right)\cdot \sin\theta\) rad/sec.
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.
By the Law of Cosines \(120^2=x^2+40^2-80x\cos\theta\text{.}\) It follows that \(x=40(\cos \theta+\sqrt{8+\cos \theta})\text{.}\)
\(\ds \frac{dx}{dt}=-40\left(1+\frac{\cos \theta}{\sqrt{8+\cos^2\theta}}\right)\cdot \sin\theta\) rad/sec.