tag:blogger.com,1999:blog-2433841880619171855.post8211785579299429881..comments2022-01-25T14:55:37.149+00:00Comments on Pat'sBlog: More on a Geome-Treat with a Calculus TwistUnknownnoreply@blogger.comBlogger3125tag:blogger.com,1999:blog-2433841880619171855.post-85071802974979265752021-08-07T08:02:57.202+01:002021-08-07T08:02:57.202+01:00Nice Article! I discovered many fascinating things...Nice Article! I discovered many fascinating things from this site. It is truly ideal to give such data and Tips. We are an online <a href="https://www.myassignmenthelponline.com" rel="nofollow">myassignmenthelp</a> suppliers to Australia understudies and everywhere on the world. Our principle objective is to help understudies in finishing their scholarly assignment errands on schedule. <br />myassignmenthelphttps://www.blogger.com/profile/04778613801416213710noreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-83547751706908225912021-05-17T17:24:14.128+01:002021-05-17T17:24:14.128+01:00I can verify that this will work in general. Multi...I can verify that this will work in general. Multivariable calculus tells us that to compute the tangent vector at a point, one need only compute the partial derivatives at that point. It should be clear from the method description that the new curve intersects the old curve at the point under consideration. Two different curves that intersect at a point are tangent at that point if their tangent vector points in the same direction, so we can scale our tangent if necessary. Using partial derivatives, in addition to the sum rule for derivatives, reduces the effort to computing single-variable derivatives for two cases. In the first case, consider the tangent of c*x^(2n) at the point x=t. Computing the derivative directly yields 2n*c*x^(2n-1). Using your method, we first transform c*x^(2n) to c*x^n*t^n. Its derivative is n*c*x^(n-1)*t^n, which at the point x=t only differs from the first calculation by a factor of 2. We similarly compute the derivative of c*x^(2n+1) at the point x=t. Direct calculation yields (2n+1)*c*x^(2n). Using your method, we first transform c*x^(2n+1) into (c*x^(n+1)*t^n + c*x^n*t^(n+1))/2. Its derivative is ((n+1)*c*x^n*t^n + n*c*x^(n-1)*t^(n+1))/2, which at the point x=t only differs from the first calculation by a factor of 2. Thus we see that the tangent vector to the curve given by your transformation is the same vector as the tangent vector of the original curve, compressed by a factor of 2. This shows that the new curve is tangent to the original curve, and by induction, we are done.Thomas Morganhttps://www.blogger.com/profile/17103380839310843667noreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-89508808394575924622021-05-17T16:43:36.935+01:002021-05-17T16:43:36.935+01:00This comment has been removed by the author.Thomas Morganhttps://www.blogger.com/profile/17103380839310843667noreply@blogger.com