write the equation of the tangent line through (1,2) for
Thank you for the opportunity to help you with your question!
The answer ends up being:
:) The full derivation requires more time than this question allows to write out. Let me know if there's anything more I can help with.
can you tell me where I went wrong see below:
sqrtx/ y ^2+4 +15/8 = 2x
sqrtx/ y ^2+4 +1.87 = 2x
sqrtx/ y ^2= 2x + 5.87
sqrt x/ y ^2=2x + 5.87
y^2 = sqrtx /(2x + 5.87)
y = sqrt [sqrt(x )/ (2x + 5.87)]
by adding given point
2= sqrt [sqrt(1 )/ (2(1) + 5.87)]+c
2= sqrt [1 / 7.87]+c
equation of tangent
Y= M x+C
y= 4 x + 5.6
Arg! I had a whole response written up but clicked the back button.
Don't simplify the 15/8. With calculus, decimals are a headache to deal with. When and where possible, just keep things as fractions.
Simplify the expression by clearing denominators, so multiply every term by 8y^2:
(8y^2)(sqrtx/y^2) + (8y^2)4 + (8y^2)(15/8) = 2x(8y^2)
8sqrtx + 32y^2 + 15y^2 = 16xy^2
No more denominators! Which means we can avoid the annoying rules about deriving with denominators. Now we just need to combine like terms:
8sqrtx + 47y^2 = 16xy^2
And then isolate y:
8sqrtx = 16xy^2 - 47y^2
8sqrtx = y^2(16x - 47)
y^2 = 8sqrtx/(16x - 47)
And then the first derivative:
Solve for point (1,2) to find the slope at that point, which ends up being:
m = -63/961
And that's it! That's the hard point. :) The rest is just finding b with the slope of a line equation.
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?