3-3-51 perpendicular at a location
X
00:00
/
00:00
CC
We want to find all the points that are perpendicular to the
line, so all the points on this graph with tangent lines.
So we're really trying to find the tangent here.
So we want the derivative, and because it's a fraction, we're
going to use the quotient rule.
The derivative of the top times the bottom minus the derivative
of the bottom times the top all over the bottom squared.
So if we simplify this up, the X's in the numerator are going
to cancel and we get -3 / X -, 3 quantity squared.
Now we want the perpendicular to the line.
So we know the slope of this line is 3.
So the perpendicular slope is negative 1/3.
So I want the slope to be -1 third.
And what that really means is I want the derivative to equal
that negative 1/3.
So we're going to cross multiply and solve.
We're going to get X -, 3 quantity squared equaling 9.
The negatives on each side will cancel.
When we square root, we get X -, 3 equaling positive or -3.
So X = 3 ± 3, X equals 6, and X = 0.
So find all the points.
So when X is 6, what's our Y?
To get our Y, we're just going to substitute it back into the
original.
So we're going to get y = 6 / 6 - 3, so 6 / 3 or two.
To get our next Y, we're going to put zero in and we're going
to get zero.
So our two points should be 6, two and 00.