Question

Find the derivative of V= 1000 ( 1- t/60 ) ^2 using limits

Answer 1

is this
\[V=1000(1-\frac{t}{60})^2\]?

Answer 2

??

Answer 3

yes

Answer 4

ok so you want
\[\lim_{h\rightarrow 0}\frac{V(t+h)-V(t)}{h}\]
\[1000\lim_{h\rightarrow 0}\frac{1-\frac{t+h}{60})-(1-\frac{t}{60})}{h}\]

Answer 5

all the work is the algebra in the numerator, so lets do that separately, then divide by h

Answer 6

yes the 1000 should be behind the limit isnt it?

Answer 7

\[\lim c f(x)=c\lim f(x)\] so i can pull it right out front

Answer 8

ohhhh ok

Answer 9

i did however forget the square, so let me rigth it correctly
\[1000\lim_{h\rightarrow 0}\frac{(1-\frac{t+h}{60})^2-(1-\frac{t}{60})^2}{h}\]

Answer 10

*write

Answer 11

man this is really going to be a pain with the square, but it is just algebra. again lets focus only on the numerator

Answer 12

yea, i think i messed the squaring too ..

Answer 13

messed up*

Answer 14

\[(1-\frac{t}{60})=1-\frac{t}{30}+\frac{t^2}{3600}\]

Answer 15

\[(-120+2t)*1000/60^2\]

Answer 16

where did the t+h go?

Answer 17

\[(1-\frac{t+h}{60})^2=1-\frac{t}{30}+\frac{t^2}{3600}-\frac{h}{30}+\frac{th}{1800}+\frac{h^2}{3600}\]

Answer 18

i did the second one first

Answer 19

because it was easiest

Answer 20

so now we subtract
\[(1+\frac{t+h}{60})^2-(1-\frac{t}{60})^2=-\frac{h}{30}+\frac{th}{1800}+\frac{h^2}{3600}\]

Answer 21

now we can divide by h

Answer 22

get
\[-\frac{1}{30}+\frac{t}{1800}+\frac{h}{3600}\] let go to zero and get
\[\frac{t}{1800}-\frac{1}{30}\] multiply by 1000 and get
\[\frac{5t}{9}-\frac{100}{3}\]

Answer 23

ohh, I caught my mistake now :) thanks

Answer 24

yw