This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Exercises 3.4
⇒ x(t) = 98.1t + 981e−t/10 − 981. To ﬁnd out when the object will hit the ground, we solve x(t) = 1000 for t. Therefore, we have 1000 = 98.1t + 981e−t/10 − 981 ⇒ 98.1t + 981e−t/10 = 1981. In this equation, if we ignore the term 981e−t/10 we will ﬁnd that t ≈ 20.2. But this means that we have ignored the term similar to 981e−2 ≈ 132.8 which we see is to large to ignore. Therefore, we must try to approximate t. We will use Newton’s method on the equation f (t) = 98.1t + 981e−t/10 − 1981 = 0. (If we can ﬁnd a root to this equation, we will have found the t we want.) Newton’s method generates a sequence of approximations given by the formula tn+1 = tn − f (tn ) . f (tn ) Since f (t) = 98.1 − 98.1e−t/10 = 98.1 1 − e−t/10 , the recursive equation above becomes tn+1 = tn − tn + 10e−tn /10 − (1981/98.1) . 1 − e−tn /10 (3.10) To start the process, let t0 = 1981/98.1 ≈ 20.19368, which was the approximation we obtained when we neglected the exponential term. Then, by equation (3.10) above we have t1 = 20.19368 − ⇒ 20.19368 + 10e−2.019368 − 20.19368 1 − e−2.019368 t1 ≈ 18.663121 . To ﬁnd t2 we plug this value for t1 into equation (3.10). This gives t2 ≈ 18.643753. Continuing this process, we ﬁnd that t3 ≈ 18.643749. Since t2 and t3 agree to four decimal places, an approximation for the time it takes the object to strike the ground is t ≈ 18.6437 sec. 5. We proceed similarly to the solution of Problem 1 to get F1 = 5g, F2 = −10g 125 ...
View Full Document