EXAMPLE 2 Find the equation of the tangent plane and the normal line to the surface x2 + y2 + 2z2 = 23 at (1, 2, 3). 相关知识点: 试题来源: 解析 Theorem A Tangent Planes For the surface F(x, y, z) = k, the equation of the tangent plane at (xo, yo, zo) is VF(xo. ...
Answer to: Find the equation of the tangent line and the normal line to the curve (x^2 + y^2)^2 = \frac{25}{3}(x^2 -y^2) at the point (2,-1) By...
Find an equation of the tangent line to the ellipse at the given point. \left( {x - 1} \right)}^2 \over 9} + \left( {y - 5} \right)}^2 \over {25 = 1, \quad \left( {4,5} \right) Find an equation of the tangent line to the ellipse...
Find the equation of the tangent line of the curve at the given point. 1. y^3 = 8x^4 at (1, 2) 2. 5x - xy + y^2 = 5 at (1, 1) Find an equation of the tangent line to the curve at the given point. y = e^(5x) cos(pi*x), (0, ...
高数求切线和法线 find the equations of tangent line and normal line of the curve X的立方+Y的立方—3XY=0 at the point (2开根号下3次方,4开根号下三次方) 相关知识点: 试题来源: 解析 x^2+y^3-3*x*y=0 (x^2+y^3-3*x*y)'=0' 2*x+2*y^2*y'-(3*y'*x+3*y)=0 y'=(2*x...
结果1 题目 Find the equation of the tangent line and the equation of the normal line drawn to the curve x^( 2/3)+y^( 2/3)=5 at the point (8,1). 相关知识点: 试题来源: 解析 y-1=-12(x-8)y-1=2(x-8) 反馈 收藏 ...
Find the equation of the tangent line to: f(x)=−1x2at the point(12,−4) Equation of Tangent Line: Slope of a curve: Lety=f(x)be a curve and(x1,y1)a point on the curve. The slope of a tangent line to the curve at the point(x1,y1) ...
A tangent line touches a curve at one and only one point. The equation of the tangent line can be determined using the slope-intercept or the point-slope method. The slope-intercept equation in algebraic form is y = mx + b, where "m" is the slope of the
题目1.Find the equation of the tangent to the curve y=x2,which is parallel to the x-axis2.Find the equation of the tangent to the curve y=x2-2x which is perpendicular to the line 2y=x-13.Find the equation of the normal to the curve y=3x2-2x-1 whi...
Find the equation of line tangent to the graph of f(x)=e+ln x at x=1? 答案 e是常数,f'(x)=1/x,所以切线在x=1点的斜率是1,在x=1点的函数值是e.So the equation of the tangent line of graph at x=1 is f(x)=x+e-1相关推荐 1Find the equation of line tangent to the grap...