mariina.1319 Posted September 28, 2016 Report Share Posted September 28, 2016 The normal to the curve y=(k/x)+lnx^2, for x≠0, at the point when x=2, has the equation 3x+2y=b. Find the exact value of k. k and b are real numbers. Reply Link to post Share on other sites More sharing options...
IB Math Helper Posted October 2, 2016 Report Share Posted October 2, 2016 Since f(x)=(k/x)+ln(x^2), then f'(x)= -k/(x^2) + 2/x f'(2)= -k/4 + 1 From the normal equation at x=2, y=(-3/2)x+b/2. That is the gradient of the normal is -3/2 meaning the gradient at x=2 is 2/3. So, f'(2)=2/3 2/3 = -k/4 + 1 k/4=1/3 k=4/3 Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.