Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
verify that the given point is on the curve nd find the lines that are a) tangent and normal to the curve at the given point.
x^2cos^2y siny=0 (0,pi)
solve by implicit differentiation
 one year ago
 one year ago
verify that the given point is on the curve nd find the lines that are a) tangent and normal to the curve at the given point. x^2cos^2y siny=0 (0,pi) solve by implicit differentiation
 one year ago
 one year ago

This Question is Closed

ZeHanzBest ResponseYou've already chosen the best response.0
This is the curve, I think:\[x^2\cos ^2 y\sin y=0\] If you set x = 0, all you have left is: sin y = 0 This is true for y = k*pi (k is an integer), so it is true for y = pi, so (0, pi) is on the curve. Implicit differentiation is just a quick way to differentiate without first solving the equation for y. You do everything the normal way: use the product rule etc., just remember that wherever you see y, this is a function of x, or y(x). This means that the Chain Rule plays an important role. I.e.: (siny)' = cosy*dy/dx Implicit differentiation of your curve therefore gives:\[2x \cos ^2 y 2x \sin y \cos y \frac{ dy }{ dx }\cos y \frac{ dy }{ dx }=0\]Solve for dy/dx and replace 2xsinycosy by sin2y:\[\frac{ dy }{ dx }=\frac{ 2x \cos ^2y }{ x \sin 2y\cos y }\]Substitute x=0 and y = pi to see the slope of the tangent line in (0, pi). Once you have done that, you will also immediately see the line normal to the curve!
 one year ago

roselinBest ResponseYou've already chosen the best response.0
okay,thanks. i will now solve it.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.