Question

\(\frac{d h_{4}}{d t}=0.5 t+2\) Integrating \(=\frac{t^{2}}{4}+2 t+5\) \(\frac{d h_{b}}{d t}=t+1\) Integrating, \(h_{b}=\frac{t^{2}}{2}+t+5\)

Short answer

Answer: The original functions are \(h_{4}(t) = \frac{t^2}{4}+2t+C_{1}\) and \(h_{b}(t) = \frac{t^2}{2}+t+C_{2}\), where \(C_{1}\) and \(C_{2}\) are constants of integration.

Step-by-step solution

Integrate \(\frac{dh_{4}}{dt}\) with respect to \(t\)

We are given \(\frac{dh_{4}}{dt}=0.5t+2\) and its integral, \(h_{4}(t)=\frac{t^2}{4}+2t+C_{1}\). Note that constant of integration, \(C_{1}\), has been added to the integral. #Step 2: Integrate the first derivative of \(h_{b}(t)\)#

Integrate \(\frac{dh_{b}}{dt}\) with respect to \(t\)

We are given \(\frac{dh_{b}}{dt}=t+1\). To find \(h_{b}(t)\), we need to integrate this expression with respect to \(t\). We obtain \(h_{b}(t)=\int(t+1)dt = \frac{t^{2}}{2}+t+C_{2}\), where \(C_{2}\) is the constant of integration. #Step 3: Write down the functions \(h_{4}(t)\) and \(h_{b}(t)\)#

State the final expressions for \(h_{4}(t)\) and \(h_{b}(t)\)

Utilizing the integrals obtained in Steps 1 and 2, the original functions are: - \(h_{4}(t) = \frac{t^2}{4}+2t+C_{1}\) - \(h_{b}(t) = \frac{t^2}{2}+t+C_{2}\) However, since the constant terms \(C_{1}\) and \(C_{2}\) cannot be determined without further information, this is the most simplified form of the original functions that we can provide.