Question

If 2x-3=5(y+1) is the equation of the normal line to the graph of f(x) when x=a, find the value of f '(a). Show your work and explain your reasoning

Answer 1

ok I have to take 2x-3=5(y+1) to figure out the slope of the normal line being used.

Answer 2

slope of normal line is 2/5
slope of tangent line is -5/2
after this I don't know what else to do

Answer 3

gotta go be back in a couple of hours

Answer 4

f'(a) = slope of tangent line = -5/2

you're done, right ?

Answer 5

There are two ways you can think of this problem in my mind, since my mind is messed up and can't ever seem to get this problem quite right.

We can consider the slope of a perpendicular line as simply being shifted by 90 degrees. So we simply use tangent function to help us out. \[\Large slope = \tan(\frac{\pi}{2}+\arctan(m))\] The other way is to imagine two lines that are perpendicular. Since the y-intercept doesn't matter for lines to be perpendicular, just forget about it and parameterize your lines so that they're vectors and use the fact that the dot product of perpendicular lines is always 0.

So if your given line is y=mx+b drop the b since it's not important and parameterize it to be the vector < t , mt > and then your perpendicular line will have slope "s" so it's vector will be < t , st >. Then this gives us: \[\Large t^2+mst^2=0\] Solving for our slope, we get \[\Large s=\frac{-1}{m}\] which sounds about right and works when we do a sanity check for the slope of 1 we get -1.