Equations #10 - Questions and Answers

Question 1 A juggler throws a bowling pin straight up with an initial speed of 8.20   m/s 8.20m/s. How much time elapses until the bowling pin returns to the juggler's hand? Answer The equation for a particle under constant acceleration also applies to freely falling bodies. The constant acceleration of the freely falling body is called the acceleration due to gravity: 𝑎 𝑦 = 𝑔 = − 9.8   m/s 2 a y ​ =g=−9.8m/s 2 Note: the starting point is the same as the ending point, so 𝑦 𝑓 − 𝑦 𝑖 = 0 y f ​ −y i ​ =0. To calculate the time, we will use the following equation: 𝑦 𝑓 = 𝑦 𝑖 + 𝑣 𝑦 𝑖 𝑡 + 1 2 𝑎 𝑦 𝑡 2 y f ​ =y i ​ +v y i ​ ​ t+ 2 1 ​ a y ​ t 2 𝑦 𝑓 − 𝑦 𝑖 = 𝑣 𝑦 𝑖 𝑡 + 1 2 𝑎 𝑦 𝑡 2 y f ​ −y i ​ =v y i ​ ​ t+ 2 1 ​ a y ​ t 2 0 = 8.20 𝑡 + 1 2 ( − 9.8 ) 𝑡 2 0=8.20t+ 2 1 ​ (−9.8)t 2 𝑡 = 1.67   s t=1.67s Question 2 A partial sum of an arithmetic sequence is given. Find the sum: 1 + 5 + 9 + ⋯ + 401 1+5+9+⋯+401. Answer Find the common difference 𝑑 d by subtracting two consecutive terms: 5 − 1 = 4 ,   𝑑 = 4 5−1=4,d=4 Find 𝑛 n by using the 𝑛 n-th term formula: 𝑎 𝑛 = 𝑎 1 + ( 𝑛 − 1 ) 𝑑 a n ​ =a 1 ​ +(n−1)d 401 = 1 + ( 𝑛 − 1 ) ( 4 ) 401=1+(n−1)(4) 401 = 4 𝑛 − 3 401=4n−3 4 𝑛 = 404 4n=404 𝑛 = 101 n=101 Now we can find the sum of the arithmetic sequence: 𝑆 𝑛 = 𝑛 2 ( 𝑎 1 + 𝑎 𝑛 ) S n ​ = 2 n ​ (a 1 ​ +a n ​ ) 𝑆 101 = 101 2 ( 1 + 401 ) S 101 ​ = 2 101 ​ (1+401) 𝑆 101 = 101 2 ( 402 ) S 101 ​ = 2 101 ​ (402) 𝑆 101 = 101 ( 201 ) S 101 ​ =101(201) 𝑆 101 = 20301 S 101 ​ =20301 Result: 𝑆 𝑛 = 20301 S n ​ =20301 Question 3 Identify the curve by finding a Cartesian equation for the curve: 𝑟 2 = 5 r 2 =5. Answer Due to 𝑟 2 = 5 r 2 =5: Recall that 𝑟 2 = 𝑥 2 + 𝑦 2 r 2 =x 2 +y 2 . Therefore, we have: 𝑥 2 + 𝑦 2 = 5 x 2 +y 2 =5 This could also be written as: ( 𝑥 − 0 ) 2 + ( 𝑦 − 0 ) 2 = 5 2 (x−0) 2 +(y−0) 2 = 5 ​ 2 An equation for a circle with a center at ( 0 , 0 ) (0,0) and radius 5 5 ​ is given here. Result: That is a circle's equation with a radius 5 5 ​ and center at ( 0 , 0 ) (0,0). Question 4 If 𝑥 𝑦 + 3 𝑒 𝑦 = 3 𝑒 xy+3e y =3e, find the value of 𝑦 y at the point where 𝑥 = 0 x=0. Answer 𝑥 𝑦 + 3 𝑒 𝑦 = 3 𝑒 ( 1 ) xy+3e y =3e(1) Let 𝑥 = 0 x=0: 0 ⋅ 𝑦 + 3 𝑒 𝑦 = 3 𝑒 0⋅y+3e y =3e 𝑒 𝑦 = 𝑒 e y =e 𝑦 = 1 y=1 Now, differentiate both sides of (1): 𝑑 𝑑 𝑥 ( 𝑥 𝑦 ) + 𝑑 𝑑 𝑥 ( 3 𝑒 𝑦 ) = 𝑑 𝑑 𝑥 ( 3 𝑒 ) dx d ​ (xy)+ dx d ​ (3e y )= dx d ​ (3e) 𝑥 𝑑 𝑦 𝑑 𝑥 + 𝑦 + 3 𝑒 𝑦 𝑑 𝑦 𝑑 𝑥 = 0 x dx dy ​ +y+3e y dx dy ​ =0 ( 𝑥 + 3 𝑒 𝑦 ) 𝑑 𝑦 𝑑 𝑥 + 𝑦 = 0 (x+3e y ) dx dy ​ +y=0 At 𝑥 = 0 , 𝑦 = 1 x=0,y=1: ( 0 + 3 𝑒 ) 𝑑 𝑦 𝑑 𝑥 + 1 = 0 (0+3e) dx dy ​ +1=0 3 𝑒 𝑑 𝑦 𝑑 𝑥 = − 1 3e dx dy ​ =−1 𝑑 𝑦 𝑑 𝑥 = − 1 3 𝑒 dx dy ​ =− 3e 1 ​ Question 5 Is there a number that is exactly 1 more than its cube? Answer Let 𝑥 x be a number equal to 1 more than its cube: 𝑥 = 𝑥 3 + 1 x=x 3 +1 Rearranging: 𝑥 3 − 𝑥 + 1 = 0 x 3 −x+1=0 This is the equation for a number whose cube is one less than the numerator. Try plugging in different values for 𝑥 x to get outputs above and below 𝑓 ( 𝑥 ) = 0 f(x)=0. For example: 𝑓 ( − 2 ) = ( − 2 ) 3 − ( − 2 ) + 1 = − 8 + 2 + 1 = − 5 f(−2)=(−2) 3 −(−2)+1=−8+2+1=−5 𝑓 ( 2 ) = 2 3 − 2 + 1 = 8 − 2 + 1 = 7 f(2)=2 3 −2+1=8−2+1=7 By the Intermediate Value Theorem, since 𝑓 ( 𝑥 ) f(x) is continuous, there is a root in the interval [ − 2 , 2 ] [−2,2]. Therefore, there exists a number 𝑥 x that satisfies the equation.