Question

Can anyone help with this homework question? I will attach a document.

Answer 1

\[L(x)=10+0.4x+0.0001x^2\]\[x(t)=752+23t+0.5t^2\]\[t=2\]\[{dL\over dt}=?\]

Answer 2

In two years;
\[x(2)=752+23\times(2)+0.5\times(2)^2\]\[~~~~~~~~=800\]\[L(800)=10+0.4\times(800)+0.0001\times(800)^2 \]\[~~~~~~~~~~~~~=394 \]

Answer 3

\[{dL\over dx}=0.4+0.0002x\]\[{dx \over dt}=23+t\]
\[{dL \over dt}={dL\over dx}{dx \over dt}\]\[{dL(x(t)) \over dt}=(0.4+0.0002x)(23+t) \]\[~~~~~~~~~~~~~~~~=9.20+0.4t+0.0046x+0.0002xt\]\[{dL(x(2)) \over dt}=9.20+0.4t+0.0092+0.0004t\]\[{dL(800) \over dt}=9.20+320+0.0092+0.32\]\[~~~~~~~~~~~~~~~=329.5292\]

Answer 4

The rate of change in Carbon monoxide level in two year time will be approximately \[330(ppm/yr)\]

Answer 5

do you get the process?

Answer 6

Yea, it makes sense now that I see it written out in steps. Thank you!

Answer 7

cool