Active forum topics
- Inverse Trigo
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Eliminate the Arbitrary Constants
- Law of cosines
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Integration of 4x^2/csc^3x√sinxcosx dx
- application of minima and maxima
- Sight Distance of Vertical Parabolic Curve
New forum topics
- Inverse Trigo
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
- Integration of 4x^2/csc^3x√sinxcosx dx
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Newton's Law of Cooling
- Law of cosines
- Can you help me po to solve this?
- Eliminate the Arbitrary Constants
Recent comments
- Yes.1 month ago
- Sir what if we want to find…1 month ago
- Hello po! Question lang po…1 month 2 weeks ago
- 400000=120[14π(D2−10000)]
(…2 months 3 weeks ago - Use integration by parts for…3 months 2 weeks ago
- need answer3 months 2 weeks ago
- Yes you are absolutely right…3 months 3 weeks ago
- I think what is ask is the…3 months 3 weeks ago
- $\cos \theta = \dfrac{2}{…3 months 3 weeks ago
- Why did you use (1/SQ root 5…3 months 3 weeks ago
Re: Growth problems
$\dfrac{dP}{dt} = kP$
$\dfrac{dP}{P} = k \, dt$
$\displaystyle \int \dfrac{dP}{P} = k \int dt$
$\ln P = kt + C$
$\ln P = \ln e^{kt} + C$
$\ln P - \ln e^{kt} + C$
$C = \ln \dfrac{P}{e^{kt}}$
When t = 0, P = 2
$C = \ln \dfrac{2}{e^{0}}$
$C = \ln 2$
Hence,
$\ln 2 = \ln \dfrac{P}{e^{kt}}$
$2 = \dfrac{P}{e^{kt}}$
$P = 2e^{kt}$
When t = 2, P = 3
$3 = 2e^{2k}$
$\dfrac{3}{2} = e^{2k}$
$e^k = \left( \dfrac{3}{2} \right)^{1/2}$
Thus,
$P = 2\left( \dfrac{3}{2} \right)^{t/2}$
(a) for t = 1
$P = 2\left( \dfrac{3}{2} \right)^{1/2} = 2.4495 ~ \text{oz}$ answer
(b) for t = 10
$P = 2\left( \dfrac{3}{2} \right)^{5} = 15.1875 ~ \text{oz}$ answer
Re: Growth problems: mold grows at a rate proportional to its...
post lang ako uli sir