What is the Rate of Movement of a Glacier after 20 Days?

[thomasrules]

The movement of a certain glacier can be modeled by $$d(t)=0.01t^2+0.5t$$ where d is the distance in metres that a stake on the glaciers has moved, relative to a fixed position, t days after measurements began. Find the rate at which the glacier is moving after 20 days.

What is the question asking for?

I got an answer by using: lim_h-0 f(a+h)-f(a)/h

and I got a final answer of 8.5 but the answer is 0.9m/day in the book !

 

[neutrino]

We could help you if you showed your work.

 

[thomasrules]

$$d(t)=(0.01)t^2+0.5t, P(a,f(a))=(20,14)\\\lim_{h\rightarrow 0} \frac{(0.01(400-800h+h^2)+0.5(20+h)-14)}{h}\\<br /> \lim_{h\rightarrow 0} \frac{(0.01h^2+8.5h)}{h}=8.5<br />$$

 

[neutrino]

I think it'll be simpler if you avoid the actual numbers till the last step.
$$\lim_{\Delta t\rightarrow 0} \frac{\left(0.01\left(t + \Delta t\right)^2 + 0.5\left(t+\Delta t\right) \right) - \left(0.01t^2 + 0.5t\right)}{\Delta t}$$

Put t = 20 after you evaluate the limit.

 

[HallsofIvy]

$$d(t)=(0.01)t^2+0.5t, P(a,f(a))=(20,14)\\\lim_{h\rightarrow 0} \frac{(0.01(400-800h+h^2)+0.5(20+h)-14)}{h}\\<br /> \lim_{h\rightarrow 0} \frac{(0.01h^2+8.5h)}{h}=8.5<br />$$

You're not going to like this!

$(20+ h)^2= 400+ 2(20)h+ h^2= 400+ 40h+ h^2$, not 800h!
(You also have a "-" that should be there but that's obviously a typo since it didn't affect your final answer.)

 

[thomasrules]

LMAO hallsofivy!

Right now I'm laughing my ass off LMAO

God damnit!

thanks...lol I GOT IT

btw how do you put spaces in between tex stuff,,,...it says its \\ but it doesn't work...look at my tex