d(t)= 36/1+0.3(1.3)^t, t is greater than and equal to 0. How long does it take (to the nearest tenth) for the snowball to melt to to half its original diameter?
6 years ago
At t=T=?, d =1/2
$\frac{1}{2}=\frac{36}{1+0.3\left(1.3\right)^T}$12 =361+0.3(1.3)T
Multiply both sides by 2
$1=\frac{72}{1+0.3\left(1.3\right)^T}$1=721+0.3(1.3)T
Multiply both sides by $1+0.3\left(1.3\right)^T$1+0.3(1.3)T
$1+0.3\left(1.3\right)^T=72$1+0.3(1.3)T=72
$0.3\left(1.3\right)^T=72-1$0.3(1.3)T=72−1
$\left(1.3\right)^T=\frac{71}{0.3}$(1.3)T=710.3
$\ln\left(1.3\right)^T=\ln\left(\frac{71}{0.3}\right)$ln(1.3)T=ln(710.3 )
$T\cdot\ln\left(1.3\right)=\ln\left(\frac{71}{0.3}\right)$T·ln(1.3)=ln(710.3 )
$T=\frac{\ln\left(\frac{71}{0.3}\right)}{\ln\left(1.3\right)}$T=ln(710.3 )ln(1.3)
$T=20.8s$T=20.8s
Note: The above solution is based on initial diameter of 1 unit
However, for t=0 the initial diameter is 27.69 units
So, with this diameter meter and the above approach the answer should be 6.4s
6 years ago
Answered By Sanam A
At t=T=?, d =1/2
$\frac{1}{2}=\frac{36}{1+0.3\left(1.3\right)^T}$12 =361+0.3(1.3)T
Multiply both sides by 2
$1=\frac{72}{1+0.3\left(1.3\right)^T}$1=721+0.3(1.3)T
Multiply both sides by $1+0.3\left(1.3\right)^T$1+0.3(1.3)T
$1+0.3\left(1.3\right)^T=72$1+0.3(1.3)T=72
$0.3\left(1.3\right)^T=72-1$0.3(1.3)T=72−1
$\left(1.3\right)^T=\frac{71}{0.3}$(1.3)T=710.3
$\ln\left(1.3\right)^T=\ln\left(\frac{71}{0.3}\right)$ln(1.3)T=ln(710.3 )
$T\cdot\ln\left(1.3\right)=\ln\left(\frac{71}{0.3}\right)$T·ln(1.3)=ln(710.3 )
$T=\frac{\ln\left(\frac{71}{0.3}\right)}{\ln\left(1.3\right)}$T=ln(710.3 )ln(1.3)
$T=20.8s$T=20.8s
6 years ago
Answered By Sanam A
Note: The above solution is based on initial diameter of 1 unit
However, for t=0 the initial diameter is 27.69 units
So, with this diameter meter and the above approach the answer should be 6.4s