12 The equation of a curve isy=x^{4}-8x^{2}+5.(a) Find the derivative.(\frac{dv}{dx})_{r},ofy=x^{4}-8x^{2}+5.(b) Find the coordinates of the three turning points.You must show all your working. 相关知识点: 试题来源: 解析 **(a) 函数的导数** 函数 y = x⁴...
The equation of a curve is xy=12 and the equation of a line l is 2x+y=k, where k is a constant.In the case where k=11, find the co-ordinates of the points of intersection of l and the curve. 相关知识点: 试题来源: 解析 Points of intersection are (1.5,8) and (4,3) ...
The equation of a curve is xy=12 and the equation of a line l is 2x+y=k, where k is a constant.In the case where k=11, find the co-ordinates of the points of intersection of l and the curve.相关知识点: 试题来源: 解析 Points of intersection are (1.5,8) and (4,3) ...
(ii) Find the equation of the tangent to the curve y=x^2√ (3+x) at the point where x=1.(iii) Find the coordinates of the turning points of the curve y=x^2√ (3+x). 相关知识点: 试题来源: 解析 (i) x=1correctly substitute their\ 12(3+x)^(- 1/2)and their\ 2...
The equation of a curve is given parametrically by x= 2θ( - sin 0),y =2 (1 - c os 0)。Sho wth at = c ot A t Aθ, = , a nd at Bθ ,= 。Fi ndt heequatio ns oft hetangen ts at A a ndB. 相关知识点: 试题来源: 解析 Equation of tangen ta tA isy=x-π+4。
百度试题 结果1 题目The equation of a curve is given by y=xe^(-2x).Using your answer to part (i), find y=xe^(-2x).(i) e^(-2x)(1-2x) 相关知识点: 试题来源: 解析 (-xe^(-2x))2- (e^(-2x))4+c 反馈 收藏
1 A-LEVEL的数学题The equation of a curve is y =6/(5 - 2x)(i) Calculate the gradient of the curve at the point where x=1.(ii) A point with coordinates (x,y) moves along the curve in such a way that the rate of increase ofy has a constant value of 0.02 units per second....
【题目】T he equation of a curve is y =x^2√(3+) forx≥-3 i Find (1-y)/(xx)(i) Find the equation of the tangent to the curve =x^2√(3+x) at the point where x=1.(i) Find the coordinates of the turning points of the curve y=x^2√(3+x) ...
of概念坐标平面假设在xy-平面上给定了一条曲线(图1),方程φ(z,y)=0称为一条曲线的隐式方程,如果这条曲线上的任一点的坐标(z,y)满足它,并且满足这个方程妒(z,y)=0的任意一对数z,Y是这条曲线上一个点的坐标.显然,一条曲线由它的方程所确定,因而,我们可以说用它的方程表示的曲线.VIP中学数学教学参考...
y=(1/x)+k y'=-1/x^2 y=(1/3)x^3 y'=x^2 ∵-1/x^2*x^2=-1 ∴两条曲线的切线互相垂直