MHB 2.6.284 AP calculus Exam Lamp and Shadow related rates

karush
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yes I know this is a very common problem but likewise many ways to solve it
View attachment 9488

ok I really have a hard time with these took me 2 hours to do this
looked at some examples but some had 3 variables and 10 steps
confusing to get the ratios set up... ok my take on it is here

see if you can solve it under 5 min
 
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given $\dfrac{dx}{dt}= 4 \text{ ft/sec}$

$\dfrac{15}{x+s} = \dfrac{6}{s} \implies s = \dfrac{2x}{3} \implies \dfrac{ds}{dt} = \dfrac{2}{3} \cdot \dfrac{dx}{dt} = \dfrac{8}{3} \text{ ft/sec}$
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...
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