Asked by Joel
Bacteria are growing in a circular colony one bacterium thick. The bacteria are growing at a constant rate, thus making the area of the colony increase at a constant rate of 12 mm sq. per hour. Find an equation expressing the rate of change of the radius as a function of the radius, r, in millimeters, of the colony. Plot dr/dt as a function of r. How fast is r hanging when it equals 3 mm? Describe the way dr/dt changes with the radius of the circle.
We just started learning related rates problems like the one above but I'm totally stumped on this one.
We just started learning related rates problems like the one above but I'm totally stumped on this one.
Answers
Answered by
bobpursley
area=PI8r^2
darea/dt=PI*2 r dr/dt
you are given darea/dt=12mm^2/hr
so dr/dt= 12/(2r) mm/hr
darea/dt=PI*2 r dr/dt
you are given darea/dt=12mm^2/hr
so dr/dt= 12/(2r) mm/hr
Answered by
Joel
I'm sorry, I'm not following what you did. I get that we know dx/dt = 12mm^2/hr and we want to find dr/dt but I don't know how you linked them together as a function of r. Maybe show me more steps and how you got the derivative dr/dt. Thanks
Answered by
bobpursley
wow, you are lost.
Start with the relationship
A=PI*r^2
take the derivative with respect to time:
dA/dt=PI*2*r*dr/dt
you are given dA/dt. Solve for dr/dt
Why are you stuck with using x as a variable? The world is more generous than that.
Start with the relationship
A=PI*r^2
take the derivative with respect to time:
dA/dt=PI*2*r*dr/dt
you are given dA/dt. Solve for dr/dt
Why are you stuck with using x as a variable? The world is more generous than that.
Answered by
Joel
12mm^2/hr = r2PI dr/dt
So 12/2PIr mm^2/hr = dr/dt
which is the same as dr/dt = 6/PIr mm^2/hr
Why'd you originally say dr/dt= 12/(2r) mm/hr
So 12/2PIr mm^2/hr = dr/dt
which is the same as dr/dt = 6/PIr mm^2/hr
Why'd you originally say dr/dt= 12/(2r) mm/hr
Answered by
bobpursley
I must have dropped a PI.
Answered by
Anonymous
now all u got to do is plug in the 3 for r and u will get the answer
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