Please Help Derivatives... last problem of the day...

**^_^Engineer_Adam^_^** · December 1st 2006, 04:02 AM

Find the equation of the tangent and normal line to the graph of the equation at the indicated point.

$y = \frac{4}{x^2};(2,1)$

the answer from the back of my book is:
tangent: $y=-x+3$ ; normal $y = x - 1$
how could this be?

**ThePerfectHacker** · December 1st 2006, 04:08 AM

Originally Posted by ^_^Engineer_Adam^_^

Find the equation of the tangent and normal line to the graph of the equation at the indicated point.

$y = \frac{4}{x^2};(2,1)$

the answer from the back of my book is:
tangent: $y=-x+3$ ; normal $y = x - 1$
how could this be?

Thus,
$y'=-\frac{8}{x^3}$
$x=2$
Thus, (the derivative at that point),
$y'(2)=-\frac{8}{8}=-1$
Using slope-intercept formula,
$y-y_0=-1(x-x_0)$
$(x_0,y_0)=(2,1)$
Thus,
$y-1=-x-2$
Thus,
$y=-x-1$
This is equation of tangent line.

Thus you are correct in saying the book is wrong.

**topsquark** · December 1st 2006, 04:10 AM

Originally Posted by ^_^Engineer_Adam^_^

Find the equation of the tangent and normal line to the graph of the equation at the indicated point.

$y = \frac{4}{x^2};(2,1)$

the answer from the back of my book is:
tangent: $y=-x+3$ ; normal $y = x - 1$
how could this be?

The tangent line will be of the form $y = mx + b$ .

We can get the slope from the derivative:
$y' = -\frac{8}{x^3}$

At x = 2 this becomes: $y' = -\frac{8}{8} = -1$

So we need the intercept. At x = 2 we know that y = 1, so
$1 = -1 \cdot 2 + b$

$b = 3$

So the tangent line is $y = -x + 3$

The line normal to this will have a slope of $-\frac{1}{-1} = 1$ , so the form of the normal line will be $y = x + b'$ and passes through (2, 1), so
$1 = 2 + b'$

$b' = -1$

So the normal line is $y = x - 1$ .

(Sorry about the graph. The lines don't look perpendicular because the x and y scales are different. I haven't figured out how to fix this yet.)

-Dan

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