Welcome to our community

Be a part of something great, join today!

Rate of Change

MWR

New member
Oct 26, 2013
15
Hi,

I am working on a rate of change problem and appear stumped with my calculations.

S(T) = (-0.03T^2 + 1.6T - 13.65)^-1

S'(T) = (-1)((-0.03T^2 + 1.6T - 13.65)^-2 (-0.03 (2T) + 1.6))

S'(T) = - (-0.06T + 1.6) / (-0.03T^2 + 1.6T - 13.65)^2

S'(T) = (0.06T-1.6) / (-0.03T^2+1.6T-13.65)^2 = 0

Therefore, 0.06T – 1.6 = 0. From this, we add 1.6 to the right and then divide by 0.06T to get T= 26.67.

When does this rate = 0?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The derivative is zero when $T=\dfrac{80}{3}$ as you found. :D
 

MWR

New member
Oct 26, 2013
15
The derivative is zero when $T=\dfrac{80}{3}$ as you found. :D
Thanks, Mark.

Could you help me finish out my problem? I'm not exactly sure how.

S(T) = (-0.03T^2+1.6T-13.65)^-1

N(T) = -0.85T^2 +45.4T - 547

So the derivative is zero for S(T) when T = 26.67.

How do you then find DS/DN, S(T) with respect to N(T)?

Do we use the chain rule? Exactly how should I go about solving DS/DN?

Thanks again.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, you are right, you could use the chain rule as follows:

\(\displaystyle \frac{dS}{dN}=\frac{dS}{dT}\cdot\frac{dT}{dN}= \frac{\dfrac{dS}{dT}}{\dfrac{dN}{dT}}\)
 

MWR

New member
Oct 26, 2013
15
Yes, you are right, you could use the chain rule as follows:

\(\displaystyle \frac{dS}{dN}=\frac{dS}{dT}\cdot\frac{dT}{dN}= \frac{\dfrac{dS}{dT}}{\dfrac{dN}{dT}}\)
For DT/DN, I get -1.7t + 45.4.

So, does this mean I multiply that by zero, since that's the derivative of S(T)? Wouldn't I get zero again?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
For DT/DN, I get -1.7t + 45.4.

So, does this mean I multiply that by zero, since that's the derivative of S(T)? Wouldn't I get zero again?
You have actually computed \(\displaystyle \frac{dN}{dT}\), so you now know:

\(\displaystyle \frac{dS}{dN}=\frac{\dfrac{0.06T-1.6}{\left(-0.03T^2+1.6T-13.65 \right)^2}}{-1.7t+45.4}=\frac{0.06T-1.6}{(45.4-1.7T)\left(-0.03T^2+1.6T-13.65 \right)^2}\)
 

MWR

New member
Oct 26, 2013
15
You have actually computed \(\displaystyle \frac{dN}{dT}\), so you now know:

\(\displaystyle \frac{dS}{dN}=\frac{\dfrac{0.06T-1.6}{\left(-0.03T^2+1.6T-13.65 \right)^2}}{-1.7t+45.4}=\frac{0.06T-1.6}{(45.4-1.7T)\left(-0.03T^2+1.6T-13.65 \right)^2}\)
This makes perfect sense. Thanks so much for the clarification. :)