At each point P of a curve for which x > 0 the tangent cuts the y-axis at T, and N is the foot of the perpendicular from P to the y-axis. If T is always 1 unit below N and the curve passes through the point (1,0), find the Cartesian equation of the curve.

When faced with a problem like this, it is always a good idea to draw a rough sketch of what might be going on. With the help of a diagram, we notice that PNT is a right angled triangle with part of the tangent (PT) as hypotenuse. Now let us consider the gradient of PT, or in other words the rise over run. Looking back to our diagram, we see the gradient is TN/NP. Since TN=1 and NP=x, we deduce that the gradient of PT is 1/x. Once we recall that the gradient of the tangent is in essence dy/dx at P, we are faced with a simple differential equation:
dy/dx = 1/x
Integrating both sides we get:
y = ln|x| + k
Given the curve passes through (1,0):
0 = ln|1|+k --> k=0
Also for x>0, x=|x|. Therefore the Cartesian equation of the curve in question is: y = ln(x)