Learning Lab

Acceleration:

Velocity \(v\left(t\right)\) is the first derivative of the displacement function \(s\left(t\right)\) with respect to time . \[\begin{align*} v & =\frac{ds}{dt} \end{align*}\] And acceleration is the rate of change or first derivative of the velocity function. It follows that acceleration \(a\) is the second derivative of the displacement function: \[\begin{align*} a & =\frac{d}{dt}\left(\frac{ds}{dt}\right)\\ & =\frac{d^{2}s}{dt^{2}}. \end{align*}\]

Beam deflection:

The equation which describes the deflection \(w\) of a uniform static beam in engineering makes use of the fourth derivative:

\[ EI\frac{d^{4}w}{dx^{4}}=q(x). \]

Points of inflection:

When graphing a function the second derivative is the rate of change of the derivative function. It describes the concavity of the curve and indicates whether the slope of the function is increasing or decreasing.

• If the graph of the function is concave upward the second derivative is positive - slope increasing.
• If the graph of the function is concave downward the second derivative is negative - slope decreasing.
• At a point of inflection, where the curve changes concavity, the second derivative is zero.

Example 1

Find the \(4^{th}\text{ derivative of $f\left(x\right)=\sin\left(2x\right).$ }\)

Solution:

\[\begin{align*} f^{\prime}\left(x\right) & =2\cos(2x)\\ f^{\prime\prime}\left(x\right) & =-4\sin(2x)\\ f^{\left(3\right)}(x) & =-8\cos(2x)\\ f^{\left(4\right)}(x) & =16\sin(2x). \end{align*}\] The fourth derivative of \(f\left(x\right)=\sin\left(2x\right)\) is \(f^{\left(4\right)}(x)=16\sin(2x).\)

Example 2

If \(f\left(x\right)=3x^{4}-5x^{2}+7,\) find \(f^{\prime\prime}\left(x\right).\)

Solution:

\[\begin{align*} f^{\prime}\left(x\right) & =12x^{3}-10x\\ f^{\prime\prime}\left(x\right) & =36x^{2}-10. \end{align*}\]

Example 3

Find \(\frac{d^{2}y}{dt^{2}}\) if \(y=\ln\left(1+t^{4}\right)\)

Solution:

Using the chain rule,

\[\begin{align*} \frac{dy}{dt} & =\frac{1}{1+t^{4}}4t^{3}\\ & =\frac{4t^{3}}{1+t^{4}}. \end{align*}\] Now applying the quotient rule,

\[\begin{align*} \frac{d^{2}y}{dt^{2}} & =\frac{\left(1+t^{4}\right)\cdot12t^{2}-4t^{3}\cdot4t^{3}}{\left(1+t^{4}\right)^{2}}\\ & =\frac{12t^{2}+12t^{6}-16t^{6}}{\left(1+t^{4}\right)^{2}}\\ & =\frac{12t^{2}-4t^{6}}{\left(1+t^{4}\right)^{2}}. \end{align*}\]

Example 4

Find the \(x-\)coordinates of the points of inflection for the curve with equation: \[\begin{align*} y & =\frac{x^{4}}{12}+\frac{x^{3}}{3}-\frac{3x^{2}}{2}+3x-7. \end{align*}\]

Solution:

\[\begin{align*} y^{\prime} & =\frac{4x^{3}}{12}+\frac{3x^{2}}{3}-\frac{6x}{2}+3\\ & =\frac{x^{3}}{3}+x^{2}-3x+3\\ y^{\prime\prime} & =x^{2}+2x-3. \end{align*}\]

Points of inflection will occur where \(y^{\prime\prime}=0:\) \[\begin{align*} x^{2}+2x-3 & =0\\ (x+3)(x-1) & =0\\ x=-3,\ & x=1 \end{align*}\]

There are points of inflection at \(x=-3\) and \(x=1\) and this is confirmed by the graph of the function.