hwk9 -1 - Solutions M408C FER Day 2 Related Rates 1. Two...

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Solutions – M408C FER – Day 2 Related Rates 1. Two boats are racing with constant speed toward a finish marker, boat A sailing from the south at 13 mph and boat B approaching from the east. When equidistant from the marker, the boats are 16 miles apart and the distance between them is decreasing at the rate of 17 mph. Which boat will win the race? y B x z A y x x x x x y x z dt dz y, z x = = = = + = + = = = = = 128 128 2 256 and 16 then 16 if 17 and 16 when 2 2 2 2 2 2 2 Since boats A and B are equidistant from the finish, the boat with the faster speed will win the race (i.e., we need to find which has the greater magnitude, 222 or ). We know 13 so boat A is going 13 mph (x is decreasing at 13 mph). 22 2 2 2(16)( 17) 2( 128)( dx dy dx dt dt dt d zxy dt dd x d y dt dt dt dz dx zx dy dz dx dy dt dt yz x dt dt dt dt y dy dt =− ⎡⎤ =+⇒ =+ ⎣⎦ ⎛⎞ ⎜⎟ ⎝⎠ =−⇒ = −− = () 13) 249.84 11.04 22.627 2( 128) So boat B is going 11.04 mph. Boat A is going 13 mph, so it will win the race. ≈≈ 1
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2. If a snowball melts so that its surface area decreases at a rate of 1 cm 2 /min, find the rate at which the diameter decreases when the diameter is 24 cm. The surface area of a sphere is given by: 2 4 Sr π = We are given that the surface area decreases at a rate of 1 cm 2 /min. Therefore, the rate of change of surface area (with respect to time) is -1 cm 2 /min. Or, 2 1 cm / min dS dt =− . We want to know the rate at which the diameter decreases when the diameter is 24 cm. That is, if D is the diameter, we want dD dt when 24 D = We need an expression relating D and S and currently only have an expression that relates S and r . Notice that: or 2 D = r 2 D r = So, we can rewrite the surface area formula: 2 4 = becomes 2 4 2 D S ⎛⎞ = ⎜⎟ ⎝⎠ and after simplifying, 2 SD = Now that we have an expression relating only S and D , we differentiate both sides (with respect to time) to obtain the expression dD dt . [] 2 dd dt dt ⎡⎤ = ⎣⎦ 2 dS dD D dt dt =⋅ ⋅ We want dD dt when 2 1 cm / min dS dt , and 24 D = . Plugging in those values, () 2 1 cm / min 2 24cm dD dt −= And so, 1 cm / min 48 dD dt = 2
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3. A balloon rises at the rate of 10 feet per second from a point of the ground 100 feet from an observer. Find the rate of change of the angle of elevation of the balloon from the observer when the balloon is 100 feet above the ground.
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This note was uploaded on 05/02/2010 for the course CH 52365 taught by Professor Mcdevitt during the Spring '09 term at University of Texas.

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hwk9 -1 - Solutions M408C FER Day 2 Related Rates 1. Two...

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