MATH 12002 - CALCULUS I § 2.3 (Part 3): Rates of Change - More Examples Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 6
Examples Example Suppose I throw a watermelon straight up from a tower that is 96 feet high at a velocity of 16 feet per second. The distance of the watermelon above the ground t seconds later is s ( t ) = − 16 t 2 + 16 t + 96 feet. How long does it take for the watermelon to reach the ground and at what velocity does it strike the ground? D.L. White (Kent State University) 2 / 6
Examples Solution When the melon hits the ground, the distance above the ground is 0 feet, and so it reaches the ground at the time t when s ( t ) = 0 . Therefore, 0 = − 16 t 2 + 16 t + 96 = − 16( t 2 − t − 6) = − 16( t + 2)( t − 3) , and so t = 3 or t = − 2 . But t cannot be negative, so the melon hits the ground after t = 3 seconds. We know that the velocity at time t is given by v ( t ) = s ′ ( t ) = − 32 t + 16 feet per second. Hence the velocity of the melon when it hits the ground is the velocity at time t = 3 , which is v (3) = − 32(3) + 16 = − 96 + 16 = − 80 feet per second. D.L. White (Kent State University) 3 / 6
Examples Example A kite is flying horizontally at a constant speed of 2 feet per second at an altitude of 30 feet. How fast must the string be let out when the length of the string is 50 feet in order to maintain the altitude? Solution Think of the kite starting directly over the flier, and let x be the horizontal distance from the flier t seconds later. Also, let’s set ℓ equal to the length of the string after t seconds. Both ℓ = ℓ ( t ) and x = x ( t ) are functions of time t, and we want to find the rate of change of the length ℓ with respect to time t when ℓ = 50 , i.e., d ℓ dt when ℓ = 50 feet. Since the horizontal speed is 2 feet per second, we have that x ( t ) = 2 t. It is clear that x and ℓ are related, but how? If we can write ℓ as a function of x, then we can write ℓ as a function of t and find d ℓ dt . [Continued → ] D.L. White (Kent State University) 4 / 6
Examples Solution [continued] We have the following situation at a given time: x ✲ KITE ✑ q ✑✑✑✑✑✑ 30 ft ℓ FLIER q By the Pythagorean Theorem, x 2 + 30 2 = ℓ 2 , hence ℓ ( t ) = x ( t ) 2 + 900 . � As we saw above, x ( t ) = 2 t, so � (2 t ) 2 + 900 = � 4 t 2 + 900 . ℓ ( t ) = Therefore, 4 t 2 (4 t 2 + 900) − 1 / 2 · 8 t = ℓ ′ ( t ) = 1 √ . 4 t 2 + 900 [Continued → ] D.L. White (Kent State University) 5 / 6
Examples Solution [continued] We need ℓ ′ ( t ) at the time t when ℓ = 50 feet. Note that when ℓ = 50 feet, x 2 + 30 2 = 50 2 and so � 50 2 − 30 2 = 40 . 2 t = x = Hence when ℓ = 50 , we have t = 20 , and the string must be let out at a speed of 4 · 20 80 = 80 ℓ ′ (20) = = √ 50 = 1 . 6 feet per second. 4(20) 2 + 900 � 2500 D.L. White (Kent State University) 6 / 6
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