# Thread: implicit, and tangent lines...

1. ## implicit, and tangent lines...

ok, it wants the normal line to a curve at a point is perpendicular to the tangent line at that point.
Find the equation of the normal line to this circle at Point A.
Find the equation of the normal line to this circle at Point B.

equation of the circle...

x^2 + y^2 = 25

PointA: (4,-3) PointB: (4,3)

equation of tangent line to A....4/3x-25/3

equation of tangent line to B....-4/3x+25/3

so it wants the perindicular normal line, so i figured it would be the same as the other, except the slope for A would be -3/4x, and B would be 3/4x, but it wasnt right....can you let me know whats wrong...thanks...

mathaction

oh yeah...and this one...i dont really know where to start or how to go about doing it...some help...thanks...

Use the graph of f(x) and the tangent line shown below to answer the following.

(a) If k(x) = 1/ f(x), then
k '(2) =

(b) If g(x) = f -1(x), then
g '(5) =

2. Originally Posted by mathaction
ok, it wants the normal line to a curve at a point is perpendicular to the tangent line at that point.
Find the equation of the normal line to this circle at Point A.
Find the equation of the normal line to this circle at Point B.

equation of the circle...

x^2 + y^2 = 25

PointA: (4,-3) PointB: (4,3)

equation of tangent line to A....4/3x-25/3
they didn't ask for the tangent line, why did you find it?

the slope of the tangent line here is 4/3. thus the slope of the normal line is -3/4. now use the point-slope form:

y + 3 = (-3/4)(x - 4)

and solve for y

equation of tangent line to B....-4/3x+25/3
do the same thing for this point

3. Originally Posted by mathaction
oh yeah...and this one...i dont really know where to start or how to go about doing it...some help...thanks...

Use the graph of f(x) and the tangent line shown below to answer the following.

(a) If k(x) = 1/ f(x), then
k '(2) =
$k(x) = \frac 1{f(x)}$

$\Rightarrow k'(x) = - \frac 1{[f(x)]^2} \cdot f'(x)$ By the chain rule

now just plug in 2

(b) If g(x) = f -1(x), then
g '(5) =
use the formula: $\left( f^{-1}\right)'(x) = \frac 1{f' \left( f^{-1}(x) \right)}$

so, $g(x) = \frac 1{f'(g(x))}$

now plug in 5 and evaluate. (i hope you know how to deal with the evaluating the inverse function)