Maths instantaneous rate of change

[Alexmcom]

The equation M(t)=12-0.2t^2 is my parobola equation.


It's asking me to find the instantaneous rate of change using the first principles method at exactly 6 seconds.


SO i use the formula (f+h)-f(x)/H


I start with 12-0.2(6)^2=4.8

Next I plug in 12-0.2(6+h)(6+h)

Which equals 36+12h+h^2. I multiply all 3 which gives me -7.2-2.4h-0.2h^2


So I got this left: 12-7.2-2.4-0.2h^2 - (-4.8)/H


I eradicate 12 and negative 7.2 with -4.8 which leaves me with -2.4h-0.2h^2/h


I eradicate the H exponents and im left with -2.4-0.2h

I plug in 0 for the h exponents and im left with -2.4.

is this correct?

 

[Deleted member 4993]

The equation M(t)=12-0.2t^2 is my parobola equation.


It's asking me to find the instantaneous rate of change using the first principles method at exactly 6 seconds.


SO i use the formula (f+h)-f(x)/H


I start with 12-0.2(6)^2=4.8

Next I plug in 12-0.2(6+h)(6+h)

Which equals 36+12h+h^2. I multiply all 3 which gives me -7.2-2.4h-0.2h^2


So I got this left: 12-7.2-2.4-0.2h^2 - (-4.8)/H


I eradicate 12 and negative 7.2 with -4.8 which leaves me with -2.4h-0.2h^2/h


I eradicate the H exponents and im left with -2.4-0.2h

I plug in 0 for the h exponents and im left with -2.4.

is this correct?

Correct

 

[Bob Brown MSEE]

The equation M(t)=12-0.2t^2 is my parabola equation.
It's asking me to find the instantaneous rate of change using the first principles method at exactly 6 seconds.

If you have proven that all parabola segments centered on t have the same slope...
Then
M'(6)=(M(12)-M(0))/12
M'(6)=(-16.8-12)/12 = -2.4

 

[Alexmcom]

Thank you guys

 

[stapel]

M(t)=12-0.2t^2 is my parobola equation.

It's asking me to find the instantaneous rate of change using the first principles method at exactly 6 seconds.

SO i use the formula (f+h)-f(x)/H

I start with 12-0.2(6)^2=4.8

Next I plug in 12-0.2(6+h)(6+h)

For full credit, you'll want to state what are the relationship between "H" and (the completely different variable) "h", between "M" and "f", and between "t" and "x". "I'll just use the standard "h" for the limit formulation, and the given variables for the function.

. . . . .$\displaystyle M(t)\, =\, 12\, -\, 0.2t^2$

. . . . .$\displaystyle M(t\, +\, h)\, =\, 12\, -\, 0.2\, (t\, +\, h)^2\, =\, 12\, -\, 0.2\, (t^2\, +\, 2ht\, +\, h^2)$

. . . . .$\displaystyle M(t\, +\, h)\, -\, M(t)\, =\, \left(12\, -\, 0.2\, (t^2\, +\, 2ht\, +\, h^2)\right)\, -\, \left(12\, -\, 0.2t^2\right)\, =\, -0.4ht\, -\, 0.2h^2$

. . . . .$\displaystyle M'(t)\, =\,$$\displaystyle \displaystyle \lim_{h\, \rightarrow\, 0}\,$$\displaystyle \dfrac{M(t\, +\, h)\, -\, M(t)}{h}\, =\, \displaystyle \lim_{h\, \rightarrow\, 0}\,$$\displaystyle \dfrac{-0.4ht\, -\, 0.2h^2}{h}\, =\,$$\displaystyle \displaystyle \lim_{h\, \rightarrow\, 0}\,$$\displaystyle -0.4t\, -\, 0.2h\, =\, -0.4t$

Then evaluate at t = 6.

 

[Alexmcom]

Gracias

For full credit, you'll want to state what are the relationship between "H" and (the completely different variable) "h", between "M" and "f", and between "t" and "x". "I'll just use the standard "h" for the limit formulation, and the given variables for the function.

. . . . .$\displaystyle M(t)\, =\, 12\, -\, 0.2t^2$

. . . . .$\displaystyle M(t\, +\, h)\, =\, 12\, -\, 0.2\, (t\, +\, h)^2\, =\, 12\, -\, 0.2\, (t^2\, +\, 2ht\, +\, h^2)$

. . . . .$\displaystyle M(t\, +\, h)\, -\, M(t)\, =\, \left(12\, -\, 0.2\, (t^2\, +\, 2ht\, +\, h^2)\right)\, -\, \left(12\, -\, 0.2t^2\right)\, =\, -0.4ht\, -\, 0.2h^2$

. . . . .$\displaystyle M'(t)\, =\,$$\displaystyle \displaystyle \lim_{h\, \rightarrow\, 0}\,$$\displaystyle \dfrac{M(t\, +\, h)\, -\, M(t)}{h}\, =\, \displaystyle \lim_{h\, \rightarrow\, 0}\,$$\displaystyle \dfrac{-0.4ht\, -\, 0.2h^2}{h}\, =\,$$\displaystyle \displaystyle \lim_{h\, \rightarrow\, 0}\,$$\displaystyle -0.4t\, -\, 0.2h\, =\, -0.4t$

Then evaluate at t = 6.