Question

an 18ft ladder leans against a wall. the bottom of the ladder is 5ft from the wall at t=0 and slides away at a rate of 2 ft/sec. Find the velocity at t=3

Answer 1

unless I'm missing something, the velocity is 2ft/sec and it's constant, isn't it?

Answer 2

i have to use derivatives it is a related rates problem

Answer 3

Do you have any sample at all?

Answer 4

what do you mean sample?

Answer 5

z² = x² + y²
->y dy/dt = - x dx/dt
=> dy/dt = - x dx/dt / y

At t = 3 => x = 5 ft + 3 * 2 = 11 ft
=> y = √ (18² - 11²) = 14.25 ft
Thus dy/dt = - 11 * 2 / 14.25 = -1.54 ft/sec

Answer 6

Your lecture or some homework?

Answer 7

That's the best I can visualize! It'd be perfect if I know what are you learning!

Answer 8

oh yea i had an example from lecture.. but i couldn't figure out from my notes where one of the numbers came from

Answer 9

how can the answer be negative?

Answer 10

Could you attach, so I can make sure my solution is 100% correct!

Answer 11

im not sure how to attach but it is correct thank you

Answer 12

Negative is the direction, which the laddter is sliding down!

Answer 13

How do you know it's correct :)

Answer 14

online homework

Answer 15

So you enter the result already?

Answer 16

yes

Answer 17

Yeehh, great! I hate to give sloppy answer!

Answer 18

The online homework is very tricky, sometimes if you're miss the sign or decimal point, your answer is rejected!