Slope of a curved line I just started with calculus and I came across the slope of a curve. According to definition the slope of a curve at a point is equal to the slope of tangent at that point. Since tangent is a straight line why not just find the tan of the angle of inclination of the straight line which would give us the slope of the line and eventually the slope of the curve at that point.
 A: The purpose of calculus is to give rigorous definitions to geometric concepts such as tangent lines and areas. While we all have an intuitive understanding of what a tangent line is, if we can't formalise this intuition, then finding tangents boils down to making rough approximations. With calculus, we don't need to use any measurements to say that the graph of $y=x^3$ has a slope of exactly $6$ at the point $(\sqrt{2},\sqrt{2}^3)$.
Moreover, the 'intuitive' concept of a tangent line isn't all that convincing when you take a step back and think about it. In elementary geometry, we are taught that a tangent line only intersects the graph at a single point. But the graph of $y=\cos x$ has a tangent at $x=0$ that intersects the graph at infinitely many points!

There are other curves where drawing a tangent line by hand would be near impossible. Consider, for instance, the graph of $y=x\sin(1/x)$, where $x$ is measured in radians:

Tell me, what is the slope of the tangent line to the above curve at exactly $x=1/(40\pi)$? Without calculus, not only would you be unable to find the answer, but you might consider this question to be completely meaningless.
But with calculus, this question is completely meaningful, and we can say that the answer is precisely $-40\pi$. This is something that we can't figure out simply by extending the tangent line to touch the $x$-axis, and measuring the angle of inclination. And that's the power of calculus.
A: At a given point $P (x,y)$ you have an associated slope
$$  y'=\frac{dy}{dx} =\tan \phi$$
You refer to inclination of the straight line that you have not as yet  established.
