Find the implicit equation of the curve that goes through the point (3, 1) and whose tangent and normal lines always form with the x axis a triangle whose area is equal o the slope of the tangent line. Assume y` > 0 and y > 0.
Hint: ∫ua2−u2du=a2−u2−a∗ln∣ua+a2−u2∣+C
(sorry, I don't know how to use the math writer yet)
This is a question from an introductory differential equations class. I have absolutely no idea how to do this! I haven't really gotten anywhere yet. This is what I've done:
let f(x) denote the curve we're looking for. Then the tangent line will have equation:
yt=dxdfx+C
Normal line will have equation yn=df/dx−1x+k
Together they will form a triangle with area = df/dx, at any point on f(x). I wanted to find an expression for area in terms of df/dx, simplify it, and solve the resulting differential equation, but I can't figure out how to do this! I'm getting very frustrated, as we've never been shown a question like this in lecture, and I can't find any examples in my textbook.
Help would be very much appreciated!
Hint: ∫ua2−u2du=a2−u2−a∗ln∣ua+a2−u2∣+C
(sorry, I don't know how to use the math writer yet)
This is a question from an introductory differential equations class. I have absolutely no idea how to do this! I haven't really gotten anywhere yet. This is what I've done:
let f(x) denote the curve we're looking for. Then the tangent line will have equation:
yt=dxdfx+C
Normal line will have equation yn=df/dx−1x+k
Together they will form a triangle with area = df/dx, at any point on f(x). I wanted to find an expression for area in terms of df/dx, simplify it, and solve the resulting differential equation, but I can't figure out how to do this! I'm getting very frustrated, as we've never been shown a question like this in lecture, and I can't find any examples in my textbook.
Help would be very much appreciated!
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