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 Calculus Calculus Math Forum

 January 29th, 2018, 08:26 AM #1 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 Applied integrals find the time When Bob runs a 10km marathon he generates the power (P). $\displaystyle \displaystyle P(t) = 600e^{-\frac{t}{5000}}$ Where the time is in seconds. A sugerpience contains about 40kJ energy. How long time does it take for Bob to run a 10 km if he consumes 30 sugarcubes during the marathon. [IMG]Link to problem http://www.mediafire.com/view/u68i8n8zvmcn3hk/IMG_20180129_181555.jpg[/IMG] What do I need to do to solve for t (the time)? I tried doing this first but that didn't work out well. $\displaystyle \displaystyle P(t) = 600e^{-\frac{t}{5000}}\\ \displaystyle E_{total} = E_s \cdot t\\ \displaystyle \int P(t) \ dt = E_s \cdot t\\ \displaystyle \int 600e^{-\frac{t}{5000}} \ dt = 1200\\$ No solutions for real numbers. January 29th, 2018, 09:39 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 $\displaystyle 600\int_0^t e^{-\frac{\tau}{5000}}d\tau= \left[-3000000 e^{-\frac{\tau}{5000}}\right]_0^t= -300000\left(1- e^{\frac{t}{5000}}\right)= 1200$. $\displaystyle 1- e{\frac{t}{5000}}= -\frac{1200}{300000}= -\frac{12}{3000}= -0.004$ $\displaystyle e^{-\frac{t}{5000}}- 1= 0.004$ $\displaystyle e^{-\frac{t}{5000}}= 1.004$ $\displaystyle -\frac{t}{5000}= ln(1.004)= 0.00399$ $\displaystyle t= -20$ Is it that this is negative that bothers you? Last edited by Country Boy; January 29th, 2018 at 09:52 AM. January 29th, 2018, 10:19 AM #3 Member   Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 I get t = -1.99600107. That doesn't seem realistic that Bob ran the 10 km marathon in about 2 seconds. I think you missed a 0 in -1200 / 30 0000 should be -1200 / 3 000 000 if I'm not mistaken. Last edited by DecoratorFawn82; January 29th, 2018 at 10:22 AM. January 29th, 2018, 10:33 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 3,034 Thanks: 1621 $P = \dfrac{dW}{dt} = 600e^{-t/5000}$ $\displaystyle W = 600 \int_0^T e^{-t/5000} \, dt$ $30 \cdot 40\, kJ = 1200 \, kJ$ $\displaystyle 1200000 = 600 \int_0^T e^{-t/5000} \, dt$ $\displaystyle 2000 = \int_0^T e^{-t/5000} \, dt$ $2000 = 5000\left[1-e^{-T/5000}\right]$ $e^{-T/5000} = \dfrac{3}{5}$ $T = 5000 \log(5/3) \approx 2554 \, sec \approx 43 \, min$ reasonable ... about a 7 minute per mile pace Thanks from greg1313 and Country Boy Tags applied, find, integrals, time Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Mathsishard123 Calculus 3 April 6th, 2014 01:56 AM Mathsishard123 Calculus 1 March 26th, 2014 09:37 AM Edz2012 Calculus 1 March 25th, 2012 03:24 PM Phatossi Calculus 4 February 26th, 2012 07:31 AM Obsessed_Math Calculus 4 February 26th, 2012 03:45 AM

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