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 October 29th, 2012, 07:35 AM #1 Newbie   Joined: Oct 2008 Posts: 4 Thanks: 0 Interpretating partial derivatives of a function of two vari Hi there I have this function, which is the function for determining the body mass index (BMI): $B(m,h)= \frac{m}{h^2}$ Then I have to determine these two values: $C= \frac{\partial B}{\partial m} = \frac{1}{h^2}$ $D= -\frac{1}{100}\frac{\partial B}{\partial h} = \frac{m}{50h^3}$ And then I have to interpret these values, and I guess C is how much B changes when you go along the m-axis (which in this case is the x-axis). But is there a like more tangible meaning to this value? It seems like it, but I am just not able to spot it :/
October 29th, 2012, 11:51 AM   #2
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Re: Interpretating partial derivatives of a function of two

Quote:
 Originally Posted by lo2 Hi there I have this function, which is the function for determining the body mass index (BMI): $B(m,h)= \frac{m}{h^2}$ Then I have to determine these two values: $C= \frac{\partial B}{\partial m} = \frac{1}{h^2}$ $D= -\frac{1}{100}\frac{\partial B}{\partial h} = \frac{m}{50h^3}$ And then I have to interpret these values, and I guess C is how much B changes when you go along the m-axis (which in this case is the x-axis). But is there a like more tangible meaning to this value? It seems like it, but I am just not able to spot it :/
C is the rate at which B is changing if you change only m, D is the rate at B is changing if you change only h. More than that cannot be said without knowing what "m" and "h" mean.

 October 29th, 2012, 01:15 PM #3 Newbie   Joined: Oct 2008 Posts: 4 Thanks: 0 Re: Interpretating partial derivatives of a function of two m is weight in kilograms and h is height in meters. So the change in m, does not depend on m itself, it rather depends on h?
 October 29th, 2012, 01:23 PM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Interpretating partial derivatives of a function of two Yes, that is correct: If we write $c= \frac{1}{h^2}$ then $B= cm$ so that B is a linear function of m with slope $\frac{1}{h^2}$ so that if m changes by some amount, then B changes by $\frac{1}{h^2}$ time the change in m, not m itself.

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