## Related questions with answers

BIOMEDICAL: Blood Flow It follows from Poiseuille's Law that blood flowing through certain arteries will encounter a resistance of $R(x)=0.25(1+x)^4$, where x is the distance (in meters) from the heart. Find the instantaneous rate of change of the resistance at:

a. 0 meters.

b. 1 meter.

Solution

VerifiedFrom the product rule we have that

$\begin{equation} R'(x)= \frac{d}{dx}\left[0.25(1+x)^4\right]=\frac{1}{4}\frac{d}{dx}\left(1+x\right)^4 \end{equation}$

Now, we can use the generalized power rule

$\begin{equation} \begin{aligned} R'(x)&=\frac{1}{4}\frac{d}{dx}\left(1+x\right)^4=\frac{1}{4}(4)(1+x)^3\frac{d}{dx}(1+x)=(1+x)^3 \end{aligned} \end{equation}$

a)

At $x=0$:

$\begin{equation} R'(0)=(1+0)^3=1 \end{equation}$

Therefore, the instantaneous rate of change of the resistance at 0 meters from the heart is 1 units of resistance per meters.

a)

At $x=01$:

$\begin{equation} R'(0)=(1+1)^3=8 \end{equation}$

Therefore, the instantaneous rate of change of the resistance at 1 meters from the heart is 8 units of resistance per meters.

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