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Please answer the following questions

Preview File Edit View Go Tools Bookmarks Window Help quiz05.pdf (page 1 of 2 1. A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and balloon increasing 3 sec later? 2. Electrical power The power P (watts) of an electric circuit is related to the circuit's resistance R (ohms) and current I (amperes) by the equation P RI2 (a) How are dP/dt, dR/dt, and dl/dt related if none of P, I and R are constant? (b) How dR/dt is related to dl/dt if P is constant? 92% CI, Wed 12:58 AM Q EE Reader Mail Inbox:... hultz Mat vs...TDE1 pen Folder Inbox 001 mcr Back to Inbox F tball game reen Shot 14...9 AM een Shot Back to Inbox 14...9 AM

Explanation / Answer

1)rate = distance / time

distance = sqrt [ (65 + t)^2 + (0 + 17t)^2 ]

That is distance in terms of time
We have to find the instantaneous rate of change when t = 3.

dy/dt = (130 + 2t + 578t) / (2sqrt ((65 + t)^2 + (17t)^2)
dy/dt = (130 + 580t) / (2sqrt ((65+t)^2 + 289t^2))

y'(3) = (130 + 1740) / (2sqrt (4624 + 2601))
y'(3) = 1870 / (2sqrt (7225))
y'(3) = 1870 / (2(85))
y'(3) = 1870 / 170
y'(3) = 11

11ft / s

b)P = RI^2

P' = (R')(I^2) + (R)(I^2)'   using the product rule

P' = (R')(I^2) + 2RII' using the power rule


Now switch notation


dP/dt = I^2 dR/dt + 2RI dI/dt

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