Asked by Lisa
Use the formula for slope of a line tangent to a parametric curve to find dy/dx for the curve c(s) = (s^-1-6s, -7s^3) at the point with s=-4
Answers
Answered by
Reiny
x = s^-1 - 6s
dx/ds = -s^-2 - 6
y = -7s^-3
dy/ds = 21s^-4
dy/dx = (dy/ds) / (dx/ds)
= (-s^-2 -6)/(21s^-4)
= (-s^-2 -6)/(21s^-4) * (s^4)/s^4)
= ( -s^2 - 6s^4)/21
when s = -4
dy/dx = (-16 - 1536)/21 = -1552/21
dx/ds = -s^-2 - 6
y = -7s^-3
dy/ds = 21s^-4
dy/dx = (dy/ds) / (dx/ds)
= (-s^-2 -6)/(21s^-4)
= (-s^-2 -6)/(21s^-4) * (s^4)/s^4)
= ( -s^2 - 6s^4)/21
when s = -4
dy/dx = (-16 - 1536)/21 = -1552/21
Answered by
Steve
Oops. y = -7s^3
dy/ds = -21s^2
Make that fix, then follow the rest of the steps.
dy/ds = -21s^2
Make that fix, then follow the rest of the steps.
Answered by
Reiny
Those silly copy errors will do it every time
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