Rate of Change (Please HELP): a metal rod L metres at temperature t ᵒC

[Ineedmathshelpasap]

The length of a metal rod L metres at temperature t ᵒC is given by:

L = 0.000009 t ² - 0.0006t + 4

1. Using the concept of derivative, find the rateof change of length with respect to temperature.
   
2. Hence determine the rate of change of length inmetres/ᵒC,the temperature is 103ᵒC.
   
   
   Please help! I think I could work it out if it was a +1 but because its +4, I'm unsure what to do. Could someone please take me through it asap
   
   

 

[Deleted member 4993]

The length of a metal rod L metres at temperature t ᵒC is given by:

L = 0.000009 t ² - 0.0006t + 4

1. Using the concept of derivative, find the rateof change of length with respect to temperature.
2. Hence determine the rate of change of length inmetres/ᵒC,the temperature is 103ᵒC.
   
   
   Please help! I think I could work it out if it was a +1 but because its +4, I'm unsure what to do. Could someone please take me through it asap
   
   

Could you please show us what you would have done if the equation was:

L = 0.000009 t ² - 0.0006t + 1

 

[Steven G]

Could you please show us what you would have done if the equation was:

L = 0.000009 t ² - 0.0006t + 1

Hey, that was going to be my comment!

 

[Ineedmathshelpasap]

Could you please show us what you would have done if the equation was:

L = 0.000009 t ² - 0.0006t + 1

would he answer be

a) dL/dt = 0.000018 - 0.0006t

b) 0.06 C/m² (slight guess

Or am I just way off on both

 

[ksdhart]

would he answer be

a) dL/dt = 0.000018 - 0.0006t

b) 0.06 C/m² (slight guess

Or am I just way off on both

You're close on both, but those are not the correct answers. Remember your derivative rules. I don't know what your book calls their theorems, but I'll refer to them as I learned them. First, you can use the Addition Rule to break apart the addition and subtraction into three separate derivatives.

L = 0.000009t² - 0.0006t + 1

\(\displaystyle \frac{dL}{dt}=\frac{d}{dt}\left(0.000009t^2\right)-\frac{d}{dt}\left(0.0006t\right)+\frac{d}{dt}\left(1\right)\)

Now you can use the Constant Multiple Rule. 0.000009 and 0.0006 are just constants, so you can pull them outside of the derivative operator.

\(\displaystyle \frac{dL}{dt}=0.000009\cdot \frac{d}{dt}\left(t^2\right)-0.0006\cdot \frac{d}{dt} \left(t\right)+\frac{d}{dt}\left(1\right)\)

Now, you try continuing from here, keeping in mind that the derivative of t² will itself be a function of t.