differentiation- application on tangents/normal
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the question is:
the tangent to the curve y=p/x - x at the point x=2 passes through the point (-1, 16). find the value of p.
was thinking find the dy/dx of the given curve, and sub in x=2 for the gradient of the tangent. but got stuck afterwards. would appreciate any help
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Originally posted by donkhead333:
the question is:
the tangent to the curve y=p/x - x at the point x=2 passes through the point (-1, 16). find the value of p.
was thinking find the dy/dx of the given curve, and sub in x=2 for the gradient of the tangent. but got stuck afterwards. would appreciate any help
dy/dx = -p(x^-2) - 1
At point x = 2, y = (p/2) - 2
Equation of tangent : y = mx + c, where m is the gradient, and c is the constant
Using x = 2 and y = (p/2) - 2,
(p/2) - 2 = [-p(2^-2) - 1](2) + c
p - 4 = -p - 4 + 2c
2p = 2c
p = c
Using x = -1 and y = 16,
16 = [-p((-1)^-2) - 1](-1) + c
16 = p + 1 + c
15 = p + c
Since p = c,
15 = 2p
p = 7.5
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ah...the answer key has p=12 as an answer...but the working seems correct...could anyone help doublecheck?
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Originally posted by donkhead333:
ah...the answer key has p=12 as an answer...but the working seems correct...could anyone help doublecheck?
Sorry about the careless mistake.
dy/dx should be always (-p/4) -1, regardless of the value of x.
The correct workings for the 2nd part should be :
Using x = -1 and y = 16,
16 = [-p((-2)^-2) - 1](-1) + c
16 = p/4 + 1 + c
64 = p + 4 + 4c
60 = p + 4c
Since p = c,
60 = 5p
p = 12