Answer:
Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.
How long will it take for this population to grow to a hundred rodents? To a thousand rodents?
Step-by-step explanation:
Use the initial condition when dp/dt = 1, p = 10 to get k;
![\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}](https://tex.z-dn.net/?f=%5Cfrac%7Bdp%7D%7Bdt%7D%20%3Dkp%5E2%5C%5C%5C%5C1%3Dk%2810%29%5E2%5C%5C%5C%5Ck%3D%5Cfrac%7B1%7D%7B100%7D)
Seperate the differential equation and solve for the constant C.
![\frac{dp}{p^2}=kdt\\\\-\frac{1}{p}=kt+C\\\\\frac{1}{p}=-kt+C\\\\p=-\frac{1}{kt+C} \\\\2=-\frac{1}{0+C}\\\\-\frac{1}{2}=C\\\\p(t)=-\frac{1}{\frac{t}{100}-\frac{1}{2} }\\\\p(t)=-\frac{1}{\frac{2t-100}{200} }\\\\-\frac{200}{2t-100}](https://tex.z-dn.net/?f=%5Cfrac%7Bdp%7D%7Bp%5E2%7D%3Dkdt%5C%5C%5C%5C-%5Cfrac%7B1%7D%7Bp%7D%3Dkt%2BC%5C%5C%5C%5C%5Cfrac%7B1%7D%7Bp%7D%3D-kt%2BC%5C%5C%5C%5Cp%3D-%5Cfrac%7B1%7D%7Bkt%2BC%7D%20%5C%5C%5C%5C2%3D-%5Cfrac%7B1%7D%7B0%2BC%7D%5C%5C%5C%5C-%5Cfrac%7B1%7D%7B2%7D%3DC%5C%5C%5C%5Cp%28t%29%3D-%5Cfrac%7B1%7D%7B%5Cfrac%7Bt%7D%7B100%7D-%5Cfrac%7B1%7D%7B2%7D%20%20%7D%5C%5C%5C%5Cp%28t%29%3D-%5Cfrac%7B1%7D%7B%5Cfrac%7B2t-100%7D%7B200%7D%20%7D%5C%5C%5C%5C-%5Cfrac%7B200%7D%7B2t-100%7D)
You have 100 rodents when:
![100=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{100} \\\\2t=98\\\\t=49\ months](https://tex.z-dn.net/?f=100%3D-%5Cfrac%7B200%7D%7B2t-100%7D%20%5C%5C%5C%5C2t-100%3D-%5Cfrac%7B200%7D%7B100%7D%20%5C%5C%5C%5C2t%3D98%5C%5C%5C%5Ct%3D49%5C%20months)
You have 1000 rodents when:
![1000=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{1000} \\\\2t=99.8\\\\t=49.9\ months](https://tex.z-dn.net/?f=1000%3D-%5Cfrac%7B200%7D%7B2t-100%7D%20%5C%5C%5C%5C2t-100%3D-%5Cfrac%7B200%7D%7B1000%7D%20%5C%5C%5C%5C2t%3D99.8%5C%5C%5C%5Ct%3D49.9%5C%20months)
Answer:
![\boxed{\sf{4}}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Csf%7B4%7D%7D%7D)
Step-by-step explanation:
Use a slope form.
<u>Slope:</u>
![\sf{\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{rise}{run} }](https://tex.z-dn.net/?f=%5Csf%7B%5Cdfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%3D%5Cdfrac%7Brise%7D%7Brun%7D%20%20%7D)
y₂ = (17)
y₁ = (1)
x₂ = (2)
x₁ = (-2)
Then, rewrite the problem.
Solve.
![\sf{\dfrac{17-1}{2-(-2)}=\dfrac{16}{4}=\dfrac{16\div4}{4\div4}=\dfrac{4}{1}=\boxed{\sf{4} }](https://tex.z-dn.net/?f=%5Csf%7B%5Cdfrac%7B17-1%7D%7B2-%28-2%29%7D%3D%5Cdfrac%7B16%7D%7B4%7D%3D%5Cdfrac%7B16%5Cdiv4%7D%7B4%5Cdiv4%7D%3D%5Cdfrac%7B4%7D%7B1%7D%3D%5Cboxed%7B%5Csf%7B4%7D%20%20%20%20%7D)
Therefore, the slope is 4, which is our answer.
I hope this helps you! Let me know if my answer is wrong or not.
Answer:
In mathematics, a differential equation is an equation that relates one or more functions and their derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two.
Step-by-step explanation:
Answer:
(-1,-5)
Step-by-step explanation:
Solve by substitution