MAT 137 Tutorial #6 Related rates June 12-13, 2017

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1 MAT 137 Tutorial #6 Related rates June 12-13, If a (spherical) snowball melts so that its surface area decreases at a rate of 1cm 2 /min, find the rate at which the diameter decreases when the diameter is 10cm. 2. Donke is fling at an altitude of 2 km towards a point directl over Shrek who is observing him. Donke s speed is 50 km/h. At what rate is the angle of elevation changing when the angle is π 6? 3. At midnight, Cinderella has to run awa from the ball. She runs along a straight path towards the Castle s wall at a speed of 4.0 m/s. Noticing this, the prince asks the guards to shine a spotlight on Cinderella as she begins to run. The spotlight is located on the ground m from the wall. Cinderella is 1.5 m tall and weighs 50 kg. At which rate is Cinderella s shadow on the wall changing when she is 20 m from the wall? Is the shadow increasing or decreasing? 4. [Harder question] (a) Prince Charming and Lord Farquaad decided to set radar stations to track a ship carring a ver valuable treasure. Prince Charming is located at Kingdom A and Lord Farquaad is at Kingdom B. Kingdom B is located 6 km east of Kingdom A. At a certain instant, the ship is 5 km from Kingdom A and also 5 km from Kingdom B. Draw a sketch and find out where the ship is located. (Warning! There are two possible answers.) (b) Pick one of the two answers to the previous question for this second part. At the same instant, the station in Kingdom A reads that the distance between Kingdom A and the ship is increasing at the rate of 28 km/h. The radar in Kingdom B reads that the distance between Kingdom B and the ship is increasing at 4 km/h. How fast and in what direction is the ship moving?

2 MAT 137 Tutorial #6 Related rates June 12-13, 2017 Solutions 1. The surface area of the snowball is A = 4πr 2 where r is the radius of the snowball. Let D be the diameter, then D = 2r an therefore A = 4πr 2 = π(2r) 2 = πd 2. Using the chain rule: So solving for we have da = da = 2πD = 1 da 2πD We know that da = 1cm2 /min, so when D = 10cm, we get = 1 2π(10cm) ( 1) cm2 /min = 1 20π So the diameter is decreasing at cm/min. 2. We first draw a diagram: Donke cm/s cm/min. 2 θ Shrek Let be the distance between the plane and the point directl over the observer and let θ be the angle of elevation. So tan(θ) = 2 = 2 cot(θ)

3 Differentiating implicitl with respect to t on both sides, we get d = 2 csc2 θ dθ. From there, we can solve: dθ = 1 2 sin2 θ d Since the plane is fling towards the point directl over the observer, we know that d = 50km/h. Thus when θ = π ( π ) 6, sin = 1 and we get 6 2 dθ ( π ) = 1 2 sin2 ( 50) = = 6.25rad/h 3. We begin b drawing a picture of the situation: 1.5 light wall Notice that in this picture we have two similar triangles: 1.5

4 Cinderella s height is 1.5m. The height of the shadow on the wall is and is a function of time t. The distance from the light to Cinderella s feet is, and is also a function of t. Cinderella s speed is d. The distance from the light to the wall is m. We want to calculate the value of d B similar triangles, 1.5 time: when = 10m and d = 4.0m/s. =. We can now differentiate implicitl with respect to [ ] d 1.5 = d [ ] 1.5 d 2 = 1 Hence, we have that when = 10 and d = 4.0m/s, m d = 1.5m 2 = 1.5m m (10m) 2 4.0m/s = 1.8m/s The derivative is negative, so the shadow is decreasing, which is what we would epect in this situation. 4. We call A and B the positions of the two Kingdoms. The distance between them is constant c = 6km. Let us call (, ) the coordinates of the ship using A as the origin. Let us call u the distance from A to the ship; radar station A measures both u and du. Let us call v the distance from B to the ship; radar station B measures both v and dv. ship u v A c- B

5 We want to calculate d du and when u = v = 5km, At that time, = 3km and = = 4km. = 28km/h and dv = 4km/h. Using Pthagoras Theorem we get = u 2 and (c ) = v 2. Then, we differentiate implicitl with respect to time on both equations and we get: 2 d 2(c ) d + 2 = 2udu + 2 = 2v dv This is a sstem with two equations and two unknowns (d/ and /). We substitute the values of,, u, v, du/, and dv/ and we get d = = 20km/h. The ship is moving NE (eactl 45 o between N and E) with a velocit of km/h.