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anonymous
 one year ago
I need a step by step tutorial on this basically i have no background on how to solve this. Would really appreciate the help.
Find the equation of the tangent line and normal line to the curve 2x^2+y^2=4 at (1, sqrt(2)).
anonymous
 one year ago
I need a step by step tutorial on this basically i have no background on how to solve this. Would really appreciate the help. Find the equation of the tangent line and normal line to the curve 2x^2+y^2=4 at (1, sqrt(2)).

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wolf1728
 one year ago
Best ResponseYou've already chosen the best response.02x^2+y^2=4 Wouldn't you first have to put this into the form y = ???

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have to isolate y first?

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.0Okay isolating y y^2=2x^2 +4 y = Square root (2x^2 +4 )

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.0valtajaros When someone asks a question then just leaves, I am not sticking around

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.0It helps if the asker sticks around so that I can ask some things

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the equation for a line passing through a point can be written as \[yy _{1}=(xx _{1})m\] where m is the slope of the line and \[y _{1}\] and \[x _{1}\] are the coordinates of the point that are known to us , in this case (1,sqrt(2)) since the point lies both on the curve and on the line, the slope at this point will be same for curve and the line. we know slope=dy/dx derive the given equation and put the values of x and y as, 1 and sqrt(2) respectively, you get the value of m put the values of m, x, and y in the general equation and you get the equation for tangent

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.0Okay, nitishdua31 seems aware of this so I will leave.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0really sorry im having internet problems thank you tho

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so first i have to get the derivative of y=sqrt(42x^2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://www.mathcentre.ac.uk/resources/uploaded/mctytannorm20091.pdf see if this can help :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks man ill have a look in to this! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes find the derivative and that gives you the value of slope

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so like sqrt(u) = du/2sqrt(u) from y = sqrt(42x^2) i get y' =(4x/2sqrt(42x^2) = 2/sqrt(42x^2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02(1)/sqrt(42(1)^2) = 2/sqrt(2) ???

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes ! so its basically sqrt(2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Attachment from Mathematica 9 may be a bit short on comments.

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.0valtajaros "really sorry im having internet problems thank you tho" ************************************************************** sorry :(
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