Question

Single Variable Calculus 18.01SC @ 36:41mins in; the professor says you can find the y-intercept when you put it into the equation of a tangent line y-y0=(-1/x^2)(x-x0).

Since f(x) = y then f(x0) = 1/x. Thus the eq. of the tangent line becomes y-(1/x) = (-1/x^2)(0-x0)

The y-intercept SHOULD come out to 2y0 but I end up getting y=2/x0.

Can someone help with this?

EDIT: This is from the triangle area problem from the video. I tried to find the area of the triangle and I get 2 when my y=2/x0 which is what the professor got as well. What am I doing wrong?

Answer 1

The equation of the tangent line is:
\[(y - y_{0}) = \left( \frac{ -1 }{x_{0}^{2}} \right) (x - x _{0})\]
To find the y-intercept, plug in x = 0 to get:
\[(y - y_{0}) = \left( \frac{ -1 }{x_{0}^{2}} \right) (0 - x _{0})\]
or
\[(y - y_{0}) = \frac{ 1 }{x_{0}} \]
Since the point (x0, y0) lies on the curve y = 1/x we know that
\[y _{0} = \frac{ 1 }{ x _{0} }\]
Substituting into the previous equation we get:
\[(y - y _{0}) = y_{0} \]
\[y = 2y _{0}\]
So the y-intercept is 2y0.
This is in fact the same thing as 2/x0 so both will give you the correct area of the triangle.

Answer 2

LOL that was so obvious and stupid of me. Thanks Zigs!