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 \(x=1?\)

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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 \(b?\)

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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 \(b=1.\) \( _\square \)