# How do I find the coordinates of the two points on the curve y=ln(x^2+1) where the gradient is 5/13

hala718 | Certified Educator

y= ln (x^2 + 1)

Let A(x1, y1)  and B(x2, y2) be twe points on the curve such that:

the gradient = 5/13

==> y' = 5/13

==> y= ln (x^2 + 1)

==> y' = 2x/(x^2 + 1) = 5/13

==> 13*2x = 5(x^2 + 1)

==> 26 x = 5x^2 + 5

==> 5x^2 -26x + 5 = 0

==> (5x -1)(x-5) = 0

==> x1= 1/5 ==> y1= ln (1/25 +1) = ln (26/25)

==> x2= 5 ==> y2= ln (26

Then the points are:

A (1/5, ln(26/25)      and   B(5, ln26)

william1941 | Student

We are given the function y = ln(x^2 +1)

We have to find the coordinates of the points where the gradient or slope is 5/13 .

To do this we first find the derivateive of y = ln(x^2 +1)

y' = 2x / (x^2 +1)

Now as we want the slope to be 5/13

Therefore y'= 5/13

=> 2x / (x^2 +1) = 5/13

=> 26x = 5x^2 +5

=> 5x^2 - 26 x +5 =0

=> 5x^2 -25x - x + 5 =0

=> 5x (x-5) -1(x-5) =0

=> (5x -1) (x-5) =0

=> x = 1/5 or x =5

At x= 1/5: y = ln (1/25 +1) = 0.039

At x= 5: y = ln (25+1) = 3.258.

Therefore the required points are (0.2, 0.039) and (5, 3.258)

neela | Student

To find the coordines of the 2 points on the curve  y = ln(x^2+1), we  find dy/dx and equate it to 5/13 and solve for x. Then the solution value of the x will be put in y = ln(x^2+1)to find the y coordinate value.

dy dx = {ln(x^2+1} = {1/(x^2+1)}(x^2+1)' = 2x/(x^2+1)

So  2x/(x^2+1)  = 5/13 and solve for x.

Multply by  13(x^2+1)

26x =5(x^2+1)

We write this as a quadratic equation:

5x^2-26x +5= 0.

(5x-1)(x-5) = 0

5x-1 = 0 . Or x - 5 =0

Or x = 1/5 . Or x= 5.

So to get yhe  corresponding y coordinates we put these  solution values of x in  y = ln(x^2+1)

When x = 1/5 , y = ln{(1/5)^2+1) = ln (26/25)

When x = 5, y = ln(5^2+1) = ln26.

Therefore the slope is 3/13 for the curve y = ln(x^2+1) at (1/5 , ln(26/25) and  (5 , ln25).