Question

Help calculus module 7 DBQ

please show work

Answer 1

1. Filling in the table is just a matter of plugging in the given
`x,y` values into the ODE
`(\mathrm dy)/(\mathrm dx)=\frac{xy}3`:

`\begin{array}cx&-1&-1&-1&0&0&0&1&1&1\\ y&1&2&3&1&2&3&1&2&3\\(\mathrm dy)/(\mathrm dx)&-\frac13&-\frac23&-1&0&0&0&\frac13&\frac23&1\end{array}`

2. I've attached what the slope field should look like. Basically, sketch a line of slope equal to the value of
`(\mathrm dy)/(\mathrm dx)` at the labeled point (these values are listed in the table).

3. This ODE is separable. We have

`(\mathrm dy)/(\mathrm dx)=\frac{xy}3\implies\frac{\mathrm dy}y=\frac x3\,\mathrm dx`

Integrating both sides gives

`\ln|y|=\frac{x^2}6+C\implies y=e^(x^2/6+C)=Ce^(x^2/6)`

With the initial condition
`f(0)=4`, we take
`x=0` and
`y=4` to solve for
`C`:

`4=Ce^0\implies C=4`

Then the particular solution is

`\boxed{y=4e^(x^2/6)}`

4. First, solve the ODE (also separable):

`(\mathrm dT)/(\mathrm dt)=k(T-38)\implies(\mathrm dT)/(T-38)=k\,\mathrm dt`

Integrating both sides gives

`\ln|T-38|=kt+C\implies T=38+Ce^(kt)`

Given that
`T(0)=75`, we can solve for
`C`:

`75=38+C\implies C=37`

Then use the other condition,
`T(30)=60`, to solve for
`k`:

`60=38+37e^(30k)\implies k=\frac1{30}\ln(22)/(37)`

Then the particular solution is

`T(t)=38+37e^{\left(\frac1{30}\ln(22)/(37)\right)t}`

Now, you want to know the temperature after an additional 30 minutes, i.e. 60 minutes after having placed the lemonade in the fridge. According to the particular solution, We have

`T(60)=38+37e^{2\ln(22)/(37)}\approx\boxed{51^\circ}`

5. You want to find
`t` such that
`T(t)=55`:

`55=38+37e^{\left(\frac1{30}\ln(22)/(37)\right)t}\implies(17)/(37)=e^{\left(\frac1{30}\ln(22)/(37)\right)t}`

`\implies t=(30\ln(17)/(37))/(\ln(22)/(37))\approx\boxed{45\,\mathrm{min}}`