Question

Calculate G(17), where dG/dt = t^−1/2 and G(3) = −3. (Round your answer to two decimal places.) G(17)=?

Answer 1

Answer:

$G(17) \approx 1.782$

Step-by-step explanation:

Let $\frac{dG}{dt} = t^{-1/2}$ and $G(3) = -3$, we proceed to find $G(t)$ by integrating $\frac{dG}{dt}$ in time:

$G(t) = \int {t^{-1/2}} \, dt$

$G(t) = \frac{t^{1/2}}{\frac{1}{2} } + C$

$G(t) = 2\cdot t^{1/2} + C$ (1)

Where $C$ is the constant of integration. If we know that $t = 3$ and $G(3) = -3$, then the constant of integration:

$-3 = 2\cdot 3^{1/2} + C$

$C = -6.464$  

The resultant function is $G(t) = 2\cdot t^{1/2} -6.464$.

Lastly, we evaluate this function at $t = 17$:

$G(17) = 2\cdot 17^{1/2}-6.464$

$G(17) \approx 1.782$