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anonymous
 5 years ago
Given \[y^{2} = x^{2}(x+7)\]
Find an equation for the tangent line at ( 3, 6).
Find an equation for the normal line at ( 3, 6).
Whats the difference? How do I find each? Confused. Help?
anonymous
 5 years ago
Given \[y^{2} = x^{2}(x+7)\] Find an equation for the tangent line at ( 3, 6). Find an equation for the normal line at ( 3, 6). Whats the difference? How do I find each? Confused. Help?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tangent line is line that touches the point on the curve normal line is line perpendicular to the tangent line to find tangent line differentiate the function to find slope slope of normal is the opposite reciprocal of that of the tangent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you help me solve the problem? Itried to work it out, but it looked complicated. :/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0equation for a line tangent is mTan(xx1)=yy1 equation for a normal line is (1/mTan)(xx1)=yy1 mtan=derivative since it touches a point once

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=\sqrt{x ^{2}(x+7)}\] differentiate using chain rule and product rule u = x^2(x+7) du = 2x(x+7) + x^2 derivitive of sqrt(u) = 1/2sqrt(u) \[y' = [x ^{2}+2x(x+7)]/2\sqrt{x ^{2}(x+7)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0slope of normal is opposite reciprocal of y' \[slopeNormal =2\sqrt{x ^{2}(x+7)}/[x ^{2}+2x(x+7)]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0plug in 3 for x y' = 5/4 slopeN = 4/5 Tangent: 6 = 5/4(3) +b b = 9/4 y = 5/4 x + 9/4 Normal: 6=4/5(3) +b b = 42/5 y = 4/5 x + 42/5
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