Slope of a Curve

Finding the slope of a curve at a point is one of two fundamental problems in calculus. This abstract concept has a variety of concrete realizations, like finding the velocity of a particle given its position and finding the rate of a reaction given the concentration as a function of time.

A tangent is a straight line that touches a curve at a single point and does not cross through it. The point where the curve and the tangent meet is called the point of tangency. We know that for a line $y=mx+c$ its slope at any point is $m$. The same applies to a curve. When we say the slope of a curve, we mean the slope of tangent to the curve at a point.

To find the slope $m$ of a curve at a particular point, we differentiate the equation of the curve. If the given curve is $y=f(x),$ we evaluate $\dfrac { dy }{ dx }$ or $f'(x)$ and substitute the value of $x$ to find the slope.

For a line of the form (or any other form) $y=mx+c,$ we can find its slope by simply taking any two values of $x,$ ${x}_{1}$ and ${x}_{2},$ and their respective $y$ values, ${y}_{1}$ and ${y}_{2}$. We find the slope by the formula $\tan \theta =\dfrac {{y}_{1}-{y}_{2}}{{x}_{1}- {x}_{2}}$. In the case of curves our approach is somewhat different. In the above case, we had $\Delta y={y}_{1}-{y}_{2}$ and $\Delta x={x}_{1}-{x}_{2}$. Now we need to find the slope of tangent to a curve at some point. To do this we again need

$$\tan \theta =\frac {{y}_{1}-{y}_{2}}{{x}_{1}- {x}_{2}},$$

but this time $\Delta y$ and $\Delta x$ tend to zero, which means the interval is very small because it is a tangent at a point.

Notice that as the colored pairs of $x_1$ and $x_2$ come closer, the tangent shifts to a point on the graph.

When this happens we replace

$$\frac { \Delta y }{ \Delta x } \quad \text{by}\quad \frac { dy }{ dx }$$

and therefore find $\frac { dy }{ dx }.$ $_\square$

What is the slope of curve $y=x^4-x^3$ at $x=1?$

---

The given curve is $y=x^4-x^3.$ Evaluating $\frac { dy }{ dx },$ we have

$$\frac{dy}{dx}=4x^3-3x^2.$$

Substituting $x=1$ into this gives

$$\frac{dy}{dx}\Big \rvert _{x=1}=4-3=1.$$

Therefore the slope of the given curve at $x=1$ is $1.$ $_\square$

If the curve $y=2x^3-bx+a$ passes through $(19, 2)$ and its slope at $x=1$ is $5,$ then what are the values of $a$ and $b?$

---

The given curve is $y=2x^3-bx+a.$ Evaluating $\frac { dy }{ dx }$, we have

$$\frac{dy}{dx}=6x^2-b.$$

Substituting $x=1$ gives

$$\frac{dy}{dx}=6-b.$$

Since the slope of the curve at $x=1$ is $5,$ we have

$$6-b=5 \Rightarrow b=1.$$

Now, substituting $(19, 2)$ into $y=2x^3-bx+a$ gives

$$19=2(2^3)-2+a \Rightarrow a=5.$$

Therefore, $a=5$ and $b=1.$ $_\square$