If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. James D. if R1 and R2 are increasing at rates of 0.3 Ω/s and 0.2 Ω/s, respectively, how fast is R changing when R1 = 60 Ω and R2 = 120 Ω? (Round your answer to three decimal places.) 2 Answers By Expert Tutors
Andre W. answered • 06/15/21 Math Class Domination! with Andre!!
Well we know the equation for the total resistance here is is called R R= 1/(R_1) + 1/(R_2) where R_1 and R_2 are 2 parts of the total resistance R of this specific circuit We also know the rate of R_1 and R_1 which you can think of as velocity and we know the values of R_1 and R_2 which you can think of as displacement Remember from kinematics that velocity is the derivative of displacement Well the same works here! The rate of the total resistance R is the derivative of R From the given equation R= (1/(R_1))+(1/(R_2)) = ((R_2)+(R_1))/((R_1)(R_2)) by rewriting the right as one term we find its derivative with respect to time (in seconds) by differentiating using the product or quotient rule of 3 terms. the 3 terms here are R_1 , R_2 , and (((R_1)+ ((R_2)))^-1) since these 3 parts make the equation of the total resistance R ----Note: to make this problem easier I moved the denominator to the numerator which is the term ((R_1)+(R_2)) by giving it a (-1) power in the numerator which makes this problem a product rule differentiation instead of a quotient rule differentiation (much tougher) Using product rule of 3 terms I get: dR/dt = (d(R_1)/dt)(R_2)(((R_1)+(R_2))^-1) + (d(R_2)/dt)(R_1)(((R_1)+(R_2))^-1) - (d((R_1)+(R_2))/dt)((R_1)+R_2)^-2)(R_1)(R_2)) and plug in all the values given to get: = 0.156 ohms per second !!
1/R = 1/R1 + 1/R2 R-1 = R1-1 + R2-1 -R-2(dR/dt) = -R1-2(dR1/dt) - R2-2(dR2/dt) dR/dt = -R2[-(dR1/dt) / R12 - (dR2/dt) / R22] When R1 = 60 and R2 = 120, 1/R = 1/60 + 1/120 = 3/120 = 1/40. So, R = 40. Therefore, dR/dt = (-1600)[-0.3/3600 - 0.2/14400] = 0.111 Ω/sec R is increasing at the rate of 0.111 ohms/sec.
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