Normal to a curve (1 Viewer)

youngsky · May 21, 2013

Hey guys help me work this one out :/

"The equation of the normal to the curve [y = x^3 - 4x] at the point (1,-3) is:"

A) y = x + 4
B) y = x - 4
C) y = -x + 2
D) y = -x - 2

include working please

Sy123 · May 21, 2013

$y = x^3 - 4x$

$y'=3x^2 - 4$

$x=1$

$y'= 3-4 = -1$

$Since$ \ \ y' \ \ $is the gradient function of the original curve$ \ \ -1 \ \ $is the gradient of the tangent at the point (1,-3)$

$Since by definition, a normal is perpendicular to the tangent, we have$

$m_N \cdot m_T = - 1$

$m_N \ = \ $gradient of normal$$

$m_T \ = \ $gradient of tangent$$

$$Hence the gradient of the normal to the curve at the specified point is$ \ \ 1$

$Using the point gradient formula$

$y-(-3) = 1(x-1)$
$y=x-4$

$\fbox{B}$

Drongoski · May 21, 2013

y'= 3x^2 - 4 = 3(1^2) - 4 = -1 at x=1

.: gradient of normal there is +1

.: Eqn of normal is: y-(-3) = 1(x-1)

i.e. y = x-4

Normal to a curve (1 Viewer)

youngsky

poof

Sy123

This too shall pass

Drongoski

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