007A Sample Final 2, Problem 5 Detailed Solution

The velocity  $V$  of the blood flow of a skier is modeled by

$V=375(R^{2}-r^{2})$

where  $R$  is the radius of the blood vessel,  $r$  is the distance of the blood flow from the center of the vessel and is a constant. Suppose the skier's blood vessel has radius  $R=0.08$  mm and that cold weather is causing the vessel to contract at a rate of  ${\frac {dR}{dt}}=-0.01$  mm per minute. How fast is the velocity of the blood changing?

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Background Information:
The equation of the tangent line to  $f(x)$  at the point  $(a,b)$  is
$y=m(x-a)+b$  where  $m=f'(a).$

Solution:

Step 1:
We use implicit differentiation to find the derivative of the given curve.
Using the product and chain rule, we get
$6x+xy'+y+2yy'=0.$
We rearrange the terms and solve for  $y'.$
Therefore,
$xy'+2yy'=-6x-y$
and
$y'={\frac {-6x-y}{x+2y}}.$

Step 2:
Therefore, the slope of the tangent line at the point  $(1,-2)$  is
${\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {-6(1)-(-2)}{1-4}}\\&&\\&=&\displaystyle {{\frac {4}{3}}.}\end{array}}$
Hence, the equation of the tangent line to the curve at the point  $(1,-2)$  is
$y={\frac {4}{3}}(x-1)-2.$

Final Answer:
$y={\frac {4}{3}}(x-1)-2.$

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