How can you find the radius of curvature of a curve at a point where the point is not on the curve. The given point (1,1/4) is not a point on the curve.
x^2+y^2=r^2 is a circle at origin (0,0). In this equation r=2. How come you got different answer with negative sign. The stated curve never passes through the point, you are finding the curvature.
In such condition u have to switch dy/dx to dx/dy as a result values of the RHS. should be reciprocal and then y comes to numerator and can be defined ,,, that's it my dear 💖👍
You just made it waaaayyyy easier, thanks alot
It's always my pleasure to enjoy ur amazing lecture sir ❤️
The second question, can just take 1/3 as constant and differentiate x^3 only.
How can you find the radius of curvature of a curve at a point where the point is not on the curve. The given point (1,1/4) is not a point on the curve.
It was amazing ❤️❤️❤️ thank you ☺️
Is the radius of curvature at any point of a circle is the radius of the circle itself?
I think the answer to number 4 is 2 units. But idk, i may be wrong. Answer this please. Thank you :)
Yeap. The curvature is same at all points of a circle, radius too
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Yes. Radius of curvature of circle at any point on the curve is radius.
how is radius negative
shouldn't you take the absolute value
x^2+y^2=r^2 is a circle at origin (0,0). In this equation r=2. How come you got different answer with negative sign. The stated curve never passes through the point, you are finding the curvature.
Please note that points you are taking are not points on the curve.
What if the value the derivative of y is undefined?😭
In such condition u have to switch dy/dx to dx/dy as a result values of the RHS. should be reciprocal and then y comes to numerator and can be defined ,,, that's it my dear 💖👍
X2+y2=4, find radius of curvature @(1,3/4)?
Can you explain 3/4 in details.
What about the second values?...u don't solve for the second values