dy/dt = 3t² + 1 and y(1) = 5. what is t when y = 3? choose 1 answer: a t = 0 b t = -3/2 c t = 3/2 d t = -1 e…

Explanation:

Step1: Integrate the derivative

We know that $\frac{dy}{dt}=3t^{2}+1$. Integrating both sides with respect to $t$, we have $y=\int(3t^{2}+1)dt$. Using the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we get $y=t^{3}+t + C$.

Step2: Find the constant C

Given $y(1) = 5$, substitute $t = 1$ and $y=5$ into $y=t^{3}+t + C$. So $5=1^{3}+1 + C$, which simplifies to $5=2 + C$. Solving for $C$, we get $C = 3$. Thus, the function is $y=t^{3}+t + 3$.

Step3: Solve for t when y = 3

Set $y = 3$ in the equation $y=t^{3}+t + 3$. We have $3=t^{3}+t + 3$. Subtracting 3 from both sides gives $t^{3}+t=0$. Factoring out $t$, we get $t(t^{2}+1)=0$. Since $t^{2}+1>0$ for all real $t$, then $t = 0$.

Answer:

A. $t = 0$