a)Nice picture!. If B is the base then after t hours AB=40t and BS=25-30t so by

Question

a)Nice picture!. If B is the base then after t hours AB=40t and BS=25-30t so by Pythagoras, (AS)^2 =(40t)^2+(25-30t)^2 =2500t^2-1500t+625 and find when this is a minimum by finding the derivative and equating to 0. You get t=0.3 which is 18 minutes. B) tan(x)=40t/(25-30t) and finding the derivative implicitly you get sec^2(x) dx/dt = {40(25-30t)+1200t}/(25-30t)^2 = 1000/(25-30t)^2 =1000/16 whet t=0.3 Also when t=0.3, (AS)^2=369, AS=sqrt(369), BS=12 so sec(x)=sqrt(369)/12. sub into 2 lines above to get dx/dt which I make 1000/41 =24.39 rad/hour. Note that x needs to be in radians for the derivative of tan(x) to be sec^2(x).

Explanation / Answer

a)Nice picture!. If B is the base then after t hours AB=40t and BS=25-30t so by Pythagoras, (AS)^2 =(40t)^2+(25-30t)^2 =2500t^2-1500t+625 and find when this is a minimum by finding the derivative and equating to 0. You get t=0.3 which is 18 minutes. B) tan(x)=40t/(25-30t) and finding the derivative implicitly you get sec^2(x) dx/dt = {40(25-30t)+1200t}/(25-30t)^2 = 1000/(25-30t)^2 =1000/16 whet t=0.3 Also when t=0.3, (AS)^2=369, AS=sqrt(369), BS=12 so sec(x)=sqrt(369)/12. sub into 2 lines above to get dx/dt which I make 1000/41 =24.39 rad/hour. Note that x needs to be in radians for the derivative of tan(x) to be sec^2(x).

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