tany = x+y: find dy/dx, pts where tangent is vertical, and

[xc630]

Hello could someone please check my work for this problem. Any response would be appreciated

Consider the relation defined by the equation tan y= x+y for x in the open interval -2pi<x<2pi

a)Find dy/dx in terms of y
b)Find the x and y coordinates of each point where the tangent line to the grpah is vertical.
c)Find (d^2y)/(dx^2)

a) d/dx tany-y =x)
sec^2y (dy/dx) - dy/dx = 1
dy/dx (sec^2y-1) = 1
dy/dx = 1/ (sec^2-1)

b) sec^2y= 1
1/ (cos^2y) = 1
cos^2y = 1
y= 0 or pi
(0,0) and (pi, 0)

c) dy/dx = 1/ (sec^2y-1)
d^2y/dx^2= (sec^2y-1)^-1
= -1 (sec^2y-1)^-2 * tan y
= -tan y/ (SQRT (sec^2y-1))

 

[soroban]

Re: tany = x+y: find dy/dx, pts where tangent is vertical, a

Hello, xc630!

A small error in part (b) . . . a larger one in (c).

Given: \(\displaystyle \,\tan y \:=\: x\,+\,y\) for \(\displaystyle (-\2pi,\,2\pi)\)

a) Find \(\displaystyle \frac{dy}{dx}\) in terms of \(\displaystyle y\)
b) Find the \(\displaystyle x\) and \(\displaystyle y\) coordinates of each point where the tangent line to the grpah is vertical.
c) Find \(\displaystyle \frac{d^2y}{dx^2}\)

\(\displaystyle (a)\;\frac{d}{dx}[\tan y\,-\,y\:=\:x]\;\;\Rightarrow\;\;\sec^2y\left(\frac{dy}{dx}\right)\,-\,\frac{dy}{dx}\:=\:1\)
\(\displaystyle \frac{dy}{dx}(\sec^2y\,-\,1) \:= \:1\;\;\Rightarrow\;\;\frac{dy}{dx}\:=\:\frac{1}{\sec^2y\,-\,1}\)

This correct, but did you notice that: \(\displaystyle \,\sec^2y\,-\,1\:=\:\tan^2y\) ?

\(\displaystyle (b)\;\sec^2y\:=\: 1\;\;\Rightarrow\;\;\frac{1}{\cos^2y}\: =\: 1\;\;\Rightarrow\;\;\cos^2y\:=\:1\;\;\Rightarrow\;\;y\:=\:0,\,\pi\;\) Right!
\(\displaystyle (0,\,0)\) and \(\displaystyle \sout{(\pi,\,0)}\;\) no

When \(\displaystyle y\,=\,\pi\), we have: \(\displaystyle \:\tan\pi\:=\:x\,+\,\pi\quad\Rightarrow\quad0 \:=\:x + \pi\quad\Rightarrow\quad x = -\pi\)

\(\displaystyle (c)\;\frac{dy}{dx}\:=\:\frac{1}{\sec^2y\,-\,1}\:=\\sec^2y\,-\,1)^{-1}\)
\(\displaystyle \frac{d^2y}{dx^2}\:=\:-1(\sec^2y\,-\,1)^{-2}\cdot\sout{\tan y}\;\) no

The derivative of \(\displaystyle \sec^2y\,-\,1\) is: \(\displaystyle \,2\cdot\sec y\,\cdot\,\sec y\cdot\tan y \:=\:2\cdot\sec^2y\cdot\tan y\)

Actually, you had: \(\displaystyle \,\frac{dy}{dx}\:=\:\frac{1}{\tan^2y}\:=\:\cot^2y\)

Then: \(\displaystyle \:\frac{d^2y}{dx^2}\:=\:-2\cdot\cot y\cdot\csc^2y\)