Gradient – Normal equation to surface with ln – Exercise 4379 Post category:Gradient Post comments:0 Comments Exercise Calculate the normal equation for the surface z=y+lnx2z=\frac{y+\ln x}{2}z=2y+lnx At point (1,1). Final Answer Show final answer x−1=y−1=z−12−2x-1=y-1=\frac{z-\frac{1}{2}}{-2}x−1=y−1=−2z−21 Solution Coming soon… Share with Friends Read more articles Previous PostGradient – A tangent plane equation for a level surface – Exercise 4382 Next PostGradient – Normal equation to surface with arctan – Exercise 4376 You Might Also Like Gradient – A scalar field with ln and a square root – Exercise 4254 March 24, 2019 Gradient – Calculate scalar field gradient and direction – Exercise 4257 March 24, 2019 Gradient – A scalar field of x multiplied by an exponential function – Exercise 4262 March 24, 2019 Gradient – Calculate maximum direction – Exercise 4265 March 24, 2019 Gradient – Calculate points where a particular gradient is obtained – Exercise 4275 March 24, 2019 Gradient – Tangent Plane Equation – Exercise 4361 March 26, 2019 Leave a Reply Cancel replyCommentEnter your name or username to comment Enter your email address to comment Enter your website URL (optional) Δ
