Differentiation Q: Do I need to differentiate to find instantaneous rates of change?

[Xibu4]

Question: Find the instantaneous rate of change in your functions at a unique time and show the average rate of change within the specified domain (28.4<=t<=64) between two unique time periods.

My functions being :
[math]f(t)=600[(1\frac{1}{3}+1)^{6t+1}-1][/math]
and

[math]g(t)= 7500e^{0.02t+0.5cos(\frac{Π}{6}t-4Π)}[/math]
Would I have to find the derivatives of these two functions or is there an alternative way ? Asking because I could only imagine how tedious it would be to differentiate both.

 

[blamocur]

I don't see how you can avoid differentiating your functions if you need instantaneous rates. But you don't need derivatives for the average rates.

 

[Steven G]

The derivative will not be as bad as you're thinking.

 

[Xibu4]

The derivative will not be as bad as you're thinking.

I got [math]3600ln(\frac{4}{3})(\frac{4}{3})^{^(6t+1)}[/math] for the first function. Subbing in my unique time gives me a gradient of 4.257 * 10^25 which doesn't match my graphs. Is my derivative correct ?

 

[Dr.Peterson]

I got [math]3600\ln\left(\frac{4}{3}\right)\left(\frac{4}{3}\right)^{6t+1}[/math] for the first function. Subbing in my unique time gives me a gradient of 4.257 * 10^25 which doesn't match my graphs. Is my derivative correct ?

For one thing. 1 1/3 + 1 is not 4/3!

 

[Xibu4]

For one thing. 1 1/3 + 1 is not 4/3!

I made a mistake with the +1 in the first function it's not meant to be there

 

[Dr.Peterson]

I made a mistake with the +1 in the first function it's not meant to be there

Then, please show the details:

I got [math]3600ln(\frac{4}{3})(\frac{4}{3})^{^(6t+1)}[/math] for the first function. Subbing in my unique time gives me a gradient of 4.257 * 10^25 which doesn't match my graphs. Is my derivative correct ?

What is your "unique time" (whatever that means), and what do your graphs show?

 

[Xibu4]

Then, please show the details:



What is your "unique time" (whatever that means), and what do your graphs show?

I just chose t to = 29.854 because it's a minimum stationary point and I can know whether I'm right or wrong. I did the first function but after finding the derivative and subbing in 29.854 I don't get 0 so I must've done something wrong.

 

[Dr.Peterson]

I just chose t to = 29.854 because it's a minimum stationary point and I can know whether I'm right or wrong. I did the first function but after finding the derivative and subbing in 29.854 I don't get 0 so I must've done something wrong.

Why do you think "29.854 [is] a minimum stationary point"? Where is the graph I asked for?

If you explain yourself and the problem more fully, you will be likely to get better help.