Find the Tangent Line to the Polar Curve: If {eq}r = f( \theta ) {/eq} is a polar curve then the slope at point {eq}(r, \theta ) {/eq} is given by: Slope {eq}(m) = \frac{ f^{'}(\theta ) \sin (\theta ) + f(\theta ) \cos (\theta )}{f...
Find the slope of the tangent to the curvey=x3−3x+2at the point whose x-coordinate is 3. View Solution Find the equation of the tangent line to the curve : y=x3−3x+5at the point (2, 7) Find the slope of the tangent to the curvey=3x2−4xat the point, whose x - co -...
We can find the equation of the tangent line by using its slope and the point of tangency on the line. Answer and Explanation: To find the slope of the tangent line to the curve {eq}\displaystyle \, \tan y + \ln (x+y)=\sin (\frac{\p...
Use implicit differentiation to find the slope of the tangent line to the curvexy3+xy=2at the point (1,1).求一个过程xy^3+xy=2 相关知识点: 试题来源: 解析 将xy3+xy=2两边对x求导 可得:y^3+3y^2y'x+y+y'x=0 将(1,1)代入可得 y'=-0.5切线方程为:y-1=-0.5(x-1) 整理得:y=-...
Answer to: Find an equation of the tangent line to the curve at the given point. y = (x - 1)/(x - 2) at the point (3, 2). By signing up, you'll get...
求导y'=6(sinx+xcosx)在π/2处y'=6 切线方程y-3π=6(x-π/2)即y=6x
Find an equation for the tangent line to the curve x^2y + xy^3 = 2 at the point (1; 1). 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 题目意思是,求曲线 x^2y + xy^3 = 2在点(1,1)处的切线方程.对方程两边取全导,d(x^2y)+d(xy^3)=2,2xydx+x^2dy+y...
The tangent to a curve is a straight line that touches the curve at a certain point and has exactly the same slope as the curve at that point. There will be a different tangent for each point of a curve, but by using calculus you will be able to calculat
结果1 题目 A curve is defined by x(t)=t^2-5t+2 and y(t)= 1((t+2)^2) for 0≤q t≤q 3. Find the equation of the tangent line to the curve when t=1. 相关知识点: 试题来源: 解析 y- 19= 2(81)(x+2) or y= 2(81)x+ (13)(81). 反馈 收藏 ...
To get a line tangent to a curve, there are two things needed -- a point and slope, which is substituted to the equation: {eq}y- y_0 = m(x-x_0) {/eq} Here: {eq}(x_0,\,y_0) {/eq} is the point where the line is tangent to the curve, and ...