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# applying differential equations to real-life problems Watch

1. (Original post by Alen.m)
So basically it's a graph of y=30000-20000e^0.2tIn(0.9)? The reason im asking is because the text book stated it's a graph of e^0.2In(0.9t) which should definetely be different Attachment 511585511587
It's sloppy notation on their part, what they mean is
2. (Original post by Zacken)
It's sloppy notation on their part, what they mean is
how about if i wanna sketch the graph of y=30000-20000e^0.2In(0.9t), in this case actually as t increases the value of e^0.2In(0.9t) increases as well therefore the value of c will decrease and the graph would look like something like this?
Attached Images

3. (Original post by Alen.m)
how about if i wanna sketch the graph of y=30000-20000e^0.2In(0.9t), in this case actually as t increases the value of e^0.2In(0.9t) increases as well therefore the value of c will decrease and the graph would look like something like this?
Not quite, if you wanted to sketch that graph, as soon as , the value for the exponential becomes positive, which makes so your whole thing becomes negative. It would look like this:

4. (Original post by Zacken)
Not quite, if you wanted to sketch that graph, as soon as , the value for the exponential becomes positive, which makes so your whole thing becomes negative. It would look like this:

thanks
5. (Original post by Zacken)
Not quite, if you wanted to sketch that graph, as soon as , the value for the exponential becomes positive, which makes so your whole thing becomes negative. It would look like this:

how about before 1.12 tho? the value of c getting smaller and smaller until t reaches 1.12 then it gets negative ?
6. (Original post by Alen.m)
how about before 1.12 tho? the value of c getting smaller and smaller until t reaches 1.12 then it gets negative ?
Yep.
7. (Original post by Zacken)
Yep.
Shouldnt it be something like this tho? Coz c start getting negative as t reaches 1.12 but on yours c is already negative before 1.12
8. (Original post by Alen.m)
Shouldnt it be something like this tho? Coz c start getting negative as t reaches 1.12 but on yours c is already negative before 1.12
That's correct. Look at the scale of my graph. :-)

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