SOLUTION: PLEASE HELP ME :) I did not know which topic to select for these questions :( 1) Find the slope of a line tangent to the curve of y= square root(3-8x)^1/3 at(-3,3). Use the deri

Question 186315: PLEASE HELP ME :) I did not know which topic to select for these questions :(
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.
2)The amount n (in g) of a compound formed during a chemical change is n=8t/[(2t^2)+3] , where t is the time (in s). Find dn/dt (derivative -I think) for t=4.0s. What is the meaning of the result???
3) Find the SECOND derivative of the function: y=square root 1-8x ???
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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.
-------------
y' = (1/2)(3-8x)^(-1/2))*(1/3)(3-8x)^(-2/3)*(-8)
-------------------------------------------------
y'(-3) = (1/2)(3+24)^(-1/2))*(1/3)(27)^(-2/3)*-8
etc.
=======================================================

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???
----
dn/dt = [[2t^2+3][8] - [8t][4t]}/ [2t^2+3]^2
----
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)]
= -4[(1-8x)^(-1/2)]
=============================
y'' = 4{(-1/2)(1-8x)^(-3/2)}
======================================
Cheers,
Stan H.

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