Did I integrate these properly?

jschlarb · #1 $Old$ January 14th, 2009, 07:15 AM

I have this question:

1. The amount a fish grows follows the differential equation
$dL/dt = 6.48e^-0.09t$ ,

with initial condition L(0) = 5, where t is measured in years and L is measured in centimetres. How much does the fish grow between ages t = 0.5 and t = 1.5?

(Give your answer to two decimal places.)

I integrated it to $L(t) = -72e^-0.09t$ and did the definite integral calculation getting ~6cm. However, when I sub in L(0), I don't get 5. Is that relevant?

galactus · #2 $Old$ January 14th, 2009, 07:34 AM

I bet you forgot your integration constant?.

$L=-72e^{-\frac{9t}{100}}+C$

$5=-72+C$

$C=77$

$L=-72e^{\frac{-9t}{100}}+77$

Now, try your t=.5 and t=1.5

jschlarb · #3 $Old$ January 14th, 2009, 07:36 AM

*facepalms*

D'OH!

thank you

Mush · #4 $Old$ January 14th, 2009, 07:41 AM

Quote:

Originally Posted by jschlarb $View Post$

I have this question:

1. The amount a fish grows follows the differential equation
$dL/dt = 6.48e^-0.09t$ ,

with initial condition L(0) = 5, where t is measured in years and L is measured in centimetres. How much does the fish grow between ages t = 0.5 and t = 1.5?

(Give your answer to two decimal places.)

I integrated it to $L(t) = -72e^-0.09t$ and did the definite integral calculation getting ~6cm. However, when I sub in L(0), I don't get 5. Is that relevant?

You are correct apart from missing out the constant of integration (that alters your answer so that L(0) = 5). But if you did a definite integral with the limits, this is not necessary anyway, and indeed the change in length is around 6.