anonymous
  • anonymous
find the eq. to the tangent line and the normal line to the curve y= x+ square root of x, (1,2)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
dy/dx= 1+1/square root of x. at (1,2) dy/dx=2=slope of tangent. and -1/2=slope of normal. equation of tangent, y-2=2(x-1). equation of normal, y-2=-1/2(x-1)
anonymous
  • anonymous
nop
anonymous
  • anonymous
if u change the square into exponential u will have (x)^1/2

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anonymous
  • anonymous
sorry dy/ dx= 1+1/2squre root of x at (1,2) slope of tangent = 3/2..now you can solve ... i just missed 1/2 i above solution
anonymous
  • anonymous
mm nop
anonymous
  • anonymous
u have dy/dx=1+1/2square root of x^-1/2
anonymous
  • anonymous
that give u that give u 5/4
anonymous
  • anonymous
if you put 1 at the place of x you wil get. 1+1/2 sruare root of 1. now 1+1/2=3/2
anonymous
  • anonymous
yea but when u take the derivative of (x)^1/2 you will get 1/2(x)1/2-1 = to 1/2 (x)-1/2... now u plug 1
anonymous
  • anonymous
sorry
anonymous
  • anonymous
1/2(x)^-1/2
anonymous
  • anonymous
now plug 1

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