2.Find the equation of the tangent to the curve y=x^2-2x which is perpendicular to the line 2y=x-13.Find the equation of the normal to the curve y=3x^2-2x-1 which is parallel to the line y=x-3 扫码下载作业帮搜索答疑一搜即得 答案解析 查看更多优质解析 解答一 举报 1:找到平行于...
结果1 题目Use implicit differentiation to find an equation of thetangent line to the curve atthe given point.2(x^2+y^2)^2=25(x^2-y^2) (3,1)(lemniscate)y0 相关知识点: 试题来源: 解析 y=-9/(13)x+(40)/(13) 反馈 收藏
Let us now consider a path in space such that the tangent to the path at any given point is parallel to the unit vector given by Eq. (2.10b), which means that t^ corresponds to a oriented unit tangent vector to the path. In other words, at any given point, the path extends along...
Graphically, this tells us that at any point (x,y) on a solution curve of the equation, the slope of the tangent line is given by the value of the function f at that point. We can outline the solution curves by using possible tangent line segments. Such a collection of tangent line ...
Answer to: Use implicit differentiation to find an equation of the tangent line to the curve at the given point. y sin 2 x = x cos 2 y, (pi / 2, pi...
Using the equation of the curve, it is possible to find the derivative, y, in order to find the slope of the tangent line at a point. For the curve, y=(−1/2)x2+1 Δy=(2)(−1/2)x Δy=(−1)x When looking at the point (1,1/2), substitute the x coordinate into ...
hard point and hard curve Geometric point interpolation method in space with tangent directional constraint Finite algorithm to find the saddle point of a quadratic function subject to linear constraints A STOCHASTIC MAXIMIN FIXED-POINT EQUATION RELATED TO GAME TREE… 带切向约束的 B样条插值B-spline cu...
A hyperbola is a type of conic section formed by the intersection of the double cone by a plane surface, but not necessarily at the centre.
We find the solution to {x″+x=0,x(0)=1,x′(0)=0 with DSolve. The graph of y=cost looks most like the first graph in toshow, corresponding to μ=1/32. exactsol=DSolve[{x″[t]+x[t]==0,x[0]==1,x′[0]==0},x[t],t] ...
The tangent line is calculated with the help of derivatives. For a function {eq}y=f(x) {/eq}, calculate the slope by taking the derivative {eq}m=f'(x) {/eq}. Then at a given point the {eq}(x_{0},y_{0}) {/eq} equation of tangent line is given by the...