10. Derivatives and Tangent Lines

The velocity is the derivative of \(x(t)\) at \(t=3\). On the \(3^\text{rd}\) line, we cancel the \(16\)'s and expand the numerator. On the \(4^\text{th}\) line we cancel the \(1\)'s in the numerator. On the \(5^\text{th}\) line we cancel the \(h\) from the denominator. \[\begin{aligned} v(3)&=x'(3)=\lim_{h\rightarrow 0} \dfrac{x(3+h)-x(3)}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{[16-(1+h)^4]-[16-(1)^4]}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{-(1+4h+6h^2+4h^3+h^4)+1}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{-(4h+6h^2+4h^3+h^4)}{h} \\ &=\lim_{h\rightarrow 0} -(4+6h+4h^2+h^3)=-4 \\ \end{aligned}\]

The slope is the derivative of \(f(x)\) at \(x=3\). On the \(3^\text{rd}\) line, we expand \((3+h)^2\) and \((3+h)^3\). On the \(4^\text{th}\) line, we cancel and combine terms in the numerator. On the \(5^\text{th}\) line we cancel the \(h\) from the denominator. \[\begin{aligned} m&=f'(3)=\lim_{h\rightarrow 0} \dfrac{f(3+h)-f(3)}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{[6(3+h)^2-(3+h)^3]-[6\cdot3^2-3^3]}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{[6(9+6h+h^2)-(27+27h+9h^2+h^3)]-[54-27]}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{54 + 36h + 6h^2 -27 - 27h -9h^2 -h^3 - 27}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{9h-3h^2 -h^3}{h} \\ &=\lim_{h\rightarrow 0} (9-3h-h^2)=9 \end{aligned}\]

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