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February 1, 2015

February 1, 2015

Posted by **=/** on Wednesday, October 5, 2011 at 12:38am.

- calculus -
**drwls**, Wednesday, October 5, 2011 at 1:01amdx/dt = 3 units/s at the time of interest

dy/dt = dy/dx * dx/dt = 3 dy/dx

dy/dx = (5/2)/sqrt(3x+1) = (5/2)/4 = 5/8

Compute dy/dt and then use

dR/dt = sqrt[(dy/dt)^2 + (dx/dt)^2] ]

sqrt[3^2 + (15/8)^2] = 3.538

- calculus -
**=/**, Wednesday, October 5, 2011 at 1:38amI just plugged it in and it says it isn't right. =/

- calculus -
**Reiny**, Wednesday, October 5, 2011 at 8:23amdrwls left out the factor of 3 in the derivative, it should have been

dy/dx = (5/2)(3) / √(3x+1)

so when x=5

dy/dx = (15/2) / √16 = 15/8

take it from there

- calculus -
**drwls**, Wednesday, October 5, 2011 at 10:35amThanks for the correction, Reiny :-)

I am getting sloppier in my old age

- calculus -
**Anonymous**, Monday, October 17, 2011 at 10:53pmat the point .

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