3.8.6-eg-1 - r ( t ? ). The only rate that we are given is...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Example A computer animator creates a circle for which the area of the circle increases at the rate of 2 mm 2 /sec. Find the rate of change of the circumference of the circle at the time that the radius is 3 mm. Solution: Let A ( t ), r ( t ), and C ( t ) denote the changing area, radius, and circumference of the circle at time t . Note that the following is given : A 0 ( t ) = 2 mm 2 / sec , which holds for all t > 0, as well as r ( t ? ) = 3 mm , where t ? > 0 is a particular instant in time. What we want is: C 0 ( t ? ) . In order to find C 0 at t = t ? , we need to relate the circumference C ( t ) to the other variables. Using the standard formula, C ( t ) = 2 πr ( t ) , which we can differentiate with respect to t to find C 0 ( t ) = 2 πr 0 ( t ) , or C 0 ( t ? ) = 2 πr 0 ( t ? ) , (1) where it remains to determine
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: r ( t ? ). The only rate that we are given is for the rate of change of area, so one idea would be to relate r ( t ) to A ( t ) in hopes of being able to learn more about r ( t ) from A ( t ). To this end, we use the area formula A ( t ) = πr 2 ( t ) , valid for t > 0, and differentiate this relationship with respect to t , A ( t ) = 2 πr ( t ) r ( t ) . It follows that for t = t ? , r ( t ? ) = A ( t ? ) 2 πr ( t ? ) = 2 2 π · 3 = 1 3 π mm/sec . Substituting this value back into (1), we determine C ( t ? ) = 2 π 1 3 π = 2 3 mm/sec ....
View Full Document

This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

Ask a homework question - tutors are online