Question 186315
1) Find the slope of a line tangent to the curve of y= square root(3-8x)^1/3 at(-3,3). Use the derivative evaluation feature of a graphing calculator to check result.
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y' = (1/2)(3-8x)^(-1/2))*(1/3)(3-8x)^(-2/3)*(-8)
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y'(-3) = (1/2)(3+24)^(-1/2))*(1/3)(27)^(-2/3)*-8
etc.
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2)The amount n (in g) of a compound formed during a chemical change is n=8t/[(2t^2)+3] , were t is the time (in s). Find dn/dt (derivative -I think) for t=4.0s. What is the meaning of the result???
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dn/dt = [[2t^2+3][8] - [8t][4t]}/ [2t^2+3]^2
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dn/dt is the rate at which n changes per t seconds
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3) Find the SECOND derivative of the function: y=square root (1-8x) ???
y' = (1/2)[(1-8x)^(-1/2)]*[-8]
= -4[(1-8x)^(-1/2)]
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y'' = 4{(-1/2)(1-8x)^(-3/2)}*[-8]
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Cheers,
Stan H.