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: 2,921 Thanks: 1518 
$P = \dfrac{dW}{dt} = 600e^{t/5000}$ $\displaystyle W = 600 \int_0^T e^{t/5000} \, dt$ $30 \cdot 40\, kJ = 1200 \, kJ$ $\displaystyle 1200000 = 600 \int_0^T e^{t/5000} \, dt$ $\displaystyle 2000 = \int_0^T e^{t/5000} \, dt$ $2000 = 5000\left[1e^{T/5000}\right]$ $e^{T/5000} = \dfrac{3}{5}$ $T = 5000 \log(5/3) \approx 2554 \, sec \approx 43 \, min$ reasonable ... about a 7 minute per mile pace 

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