10. Derivatives and Tangent Lines

c. Derivative as a Function

3. Velocity, Acceleration and Other Rates

Velocity

If we know the position as a function of time, we can also find the velocity as a function of time by simply taking the derivative.

Between\(t=0\) and \(t=4\), a girl takes a walk from \(x=0\) to \(x=16\) and back. Her position at time \(t\) is \(x(t)=16-(t-2)^4\). What is her velocity at time \(t\)? Use it to find her velocity at \(t=3\).

\(\begin{aligned} v(t)&=x'(t)=-4(t-2)^3 \\ v(3)&=x'(3)=-4 \end{aligned}\)

We find the velocity by diferentiating the position \(x(t)\).
On the \(3^\text{rd}\) line, we cancel the \(16\)'s and partially expand the numerator. (We keep powers of \(t-2\) grouped because that is what will cancel.)
On the \(4^\text{th}\) line we cancel the \((t-2)^4\)'s in the numerator.
On the \(5^\text{th}\) line we cancel the \(h\) from the denominator. \[\begin{aligned} v(t)&=x'(t)=\lim_{h\rightarrow 0} \dfrac{x(t+h)-x(t)}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{[16-(t+h-2)^4]-[16-(t-2)^4]}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{-[(t-2)^4+4(t-2)^3h+O(h^2)]+(t-2)^4}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{-4(t-2)^3h+O(h^2)}{h} \\ &=\lim_{h\rightarrow 0} (-4(t-2)^3+O(h))=-4(t-2)^3 \\ \end{aligned}\] Where \(O(h^n)\) means additional terms with powers of \(h\) of degree \(n\) or higher.

Using \(v(t)=-4(t-2)^3\), we find \(v(3)=-4\). This is the same answer we found in a previous exercise.

Acceleration

The acceleration is the rate of change of the velocity. So it can be found by differentiating the velocity.

Between\(t=0\) and \(t=4\), a girl takes a walk from \(x=0\) to \(x=16\) and back. Her position at time \(t\) is \(x(t)=16-(t-2)^4\). What is her acceleration at time \(t\)? Use it to find her acceleration at \(t=3\).

\(\begin{aligned} a(t)&=v'(t)=-12(t-2)^2 \\ a(3)&=v'(3)=-12 \end{aligned}\)

In the previous exercise, we found the velocity is: \[ v(t)=-4(t-2)^3 \] The acceleration is the derivative of this: \[\begin{aligned} a(t)&=v'(t)=\lim_{h\rightarrow 0} \dfrac{v(t+h)-v(t)}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{[-4(t+h-2)^3]-[-4(t-2)^3]}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{-4[(t-2)^3+3(t-2)^2h+3(t-2)h^2+h^3]+4(t-2)^3}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{-4[3(t-2)^2h+3(t-2)h^2+h^3]}{h} \\ &=\lim_{h\rightarrow 0} -4[3(t-2)^2+3(t-2)h+h^2]=-12(t-2)^2 \end{aligned}\]

Since \(a(t)=-12(t-2)^2\), we find \(a(3)=-12\).

Jerk

The jerk is the rate of change of the acceleration. So it can be found by differentiating the acceleration.

Between\(t=0\) and \(t=4\), a girl takes a walk from \(x=0\) to \(x=16\) and back. Her position at time \(t\) is \(x(t)=16-(t-2)^4\). What is her jerk at time \(t\)? Use it to find her jerk at \(t=3\).

\(\begin{aligned} j(t)&=a'(t)=-24(t-2) \\ j(3)&=a'(3)=-24 \end{aligned}\)

In the previous exercise, we found the acceleration is: \[ a(t)=-12(t-2)^2 \] The jerk is the derivative of this: \[\begin{aligned} j(t)&=a'(t)=\lim_{h\rightarrow 0} \dfrac{a(t+h)-a(t)}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{[-12(t+h-2)^2]-[-12(t-2)^2]}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{-12[(t-2)^2+2(t-2)h+h^2]+12(t-2)^2}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{-12[2(t-2)h+h^2]}{h} \\ &=\lim_{h\rightarrow 0} -12[2(t-2)+h]=-24(t-2) \end{aligned}\]

Since \(j(t)=-24(t-2)\), we find \(j(3)=-24\).

Other Rates of Change

The amount of water in a rain gauge at time \(t\) is given by \(R(t)=\sqrt{t}\). What is the rate at which the gauge is filling as a function of \(t\)? What is the rate at \(t=4\)?

\(\begin{aligned} R'(x)&=\dfrac{1}{2\sqrt{x}} \\ R'(4)&=\dfrac{1}{4} \end{aligned}\)

The rate at which the gauge is filling is the derivative \(R'(x)\).
On the \(3^\text{rd}\) line, we multiply the numerator and denominator by the conjugate.
On the \(4^\text{th}\) line, we expand the numerator using difference of squares.
On the \(5^\text{th}\) line we simplify the numerator and cancel the \(h\) from the denominator. \[\begin{aligned} R'(x)&=\lim_{h\rightarrow 0} \dfrac{R(x+h)-R(x)}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{\sqrt{x+h}-\sqrt{x}}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{(\sqrt{x+h}-\sqrt{x})}{h} \dfrac{(\sqrt{x+h}+\sqrt{x})}{(\sqrt{x+h}+\sqrt{x})} \\ &=\lim_{h\rightarrow 0} \dfrac{(x+h)-(x)}{h(\sqrt{x+h}+\sqrt{x})} \\ &=\lim_{h\rightarrow 0} \dfrac{1}{\sqrt{x+h}+\sqrt{x}}=\dfrac{1}{2\sqrt{x}} \\ \end{aligned}\] At \(x=4\) we have: \[ R'(4)=\dfrac{1}{2\sqrt{4}}=\dfrac{1}{4} \] Notice this is the same answer we found in a previous exercise.