# basic calc two problems! thanks for the help!

• Dec 10th 2007, 08:30 PM
wpski
basic calc two problems! thanks for the help!
ok two problems both are totally weird and have no context in my book

1) The tangent to y=(ax+b)/(sqaureroot of x) at the point where x=1 is 2x-y=1. find a and b

2) Find a given that the tangent to y=(-2)/((ax+1)^2) at x=0 passe through the point (1,0)

any help is appreciated
thank you for all your help!

ps
this is my first chapter using calc so like limits and derivatives ect.
• Dec 10th 2007, 09:40 PM
earboth
Quote:

Originally Posted by wpski
...

1) The tangent to y=(ax+b)/(sqaureroot of x) at the point where x=1 is 2x-y=1. find a and b

...

Hello,

you have:

$\displaystyle f(x)=\frac{ax+b}{x^2} = \frac ax + \frac b{x^2}$

and the equation of a tangent line:

$\displaystyle t:\ y=2x-1$

1. Calculate the coordinates of the tangent point for x = 1. Plug in this value into the equation of the tangent line: T(1, 1)
Plug in these values into the equation of the function because the coordinates of the tangent point must satisfy the equation of the function too:

$\displaystyle f(1) = 1 =\frac{a+b}{1}~\implies~\boxed{a+b=1}~\color{blue}[1]$

2. The slope of the tangent line is m = 2 and is equal to the first derivatie of f at x = 1. Calculate the first derivatie. I took the 2nd version of the term of the function:

$\displaystyle f'(x) = -\frac a{x^2} - \frac{2b}{x^3}$ . Thus

$\displaystyle f'(1) = -\frac a{1} - \frac{2b}{1}~\implies~\boxed{-a-2b = 2}~ \color{blue}[2]$

3. Solve the system of simultaneous equations for a and b. I've got a = 4 and b = -3
• Dec 10th 2007, 09:50 PM
earboth
Quote:

Originally Posted by wpski
...

2) Find a given that the tangent to y=(-2)/((ax+1)^2) at x=0 passe through the point (1,0)

...

Hi,

you have: $\displaystyle f(x) = -\frac{2}{(ax+1)^2} = -2 \cdot (ax+1)^{-2}$

The point T(0, -2) belongs to the graph of f and is the tangent point of a line which passes through the points P(1, 0) and T(0, -2).

Use the 2-point-formula to calculate the equation of the tangent line:

$\displaystyle \frac{y-0}{x-1} = \frac{-2-0}{0-1} = 2~\implies~y = 2x-2$

Thus the slope of the tangent line is m = 2

The slope of the tangent line is equal to f'(0)

$\displaystyle f'(x) = (-2) \cdot (-2) \cdot (ax+1)^{-3} \cdot a = \frac{4a}{(ax+1)^3}$ . That means $\displaystyle f'(0) = \frac{4a}{1^3}=4a$

So you get:

$\displaystyle 4a = 2~\iff~a=\frac12$