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Mathematics 19 Online
OpenStudy (anonymous):

A 14 foot ladder is leaning against a wall. If the top slips down the wall at a rate of 2 ft/s, how fast will the foot of the ladder be moving away from the wall when the top is 9 feet above the ground?

OpenStudy (anonymous):

You just want an answer huh?

OpenStudy (anonymous):

well now i do because all i have tried doesnt work

OpenStudy (anonymous):

and im fed up

OpenStudy (anonymous):

ive got all courses materials to study. i dont have time to sit around and do trials and errors

OpenStudy (anonymous):

if someone can just solve it showing steps. i will learn

OpenStudy (anonymous):

I see. Well your first step should|dw:1330481730583:dw| be drawing out the situation.

OpenStudy (anonymous):

Then you should notice that they say the side labeled b in the picture is changing at a rate of 2 m/s so that gives you the relation db/dt = 2 (m/s)

OpenStudy (anonymous):

i have got the first two steps

OpenStudy (anonymous):

Also note that the ladder does not change in length (side C...not labeled in picture) so dC/dt = 0. Now they want you to find the rate of change of a when b has a length of 9 feet. An equation that relates the sides of a right triangle is the Pythagorean theorem which is \[c^2 = b^2 +a^2\] and since the sides are changing you will have to implicitly derive this equation and you'll get \[2c {dc \over dt} = 2b{db \over dt} + 2a{da \over dt}\] and then to make things easier plug in your values for \[{dc \over dt} = 0\] and \[{db \over dt} = 2\] and since you want to find the rate of change when b is 9 feet you plug in that value for b and then solve for \[{da \over dt}\] which is the rate of change of the a side with respect to time. I hope that helps.

OpenStudy (anonymous):

ok kool thanks

OpenStudy (anonymous):

The main idea here is implicit differentiation and related rates.

OpenStudy (anonymous):

2 13.85) dx/dt + 2 (14) (2) = 0 dx/dt= 56/27.7= 2.02???

OpenStudy (anonymous):

do you agree with 2.02?

OpenStudy (anonymous):

I don't think that's right...check your length of a when b = 9 and c = 14...it should be \[\sqrt(115)\].

OpenStudy (anonymous):

Also\[ {db \over dt} = -2 {m \over s}\] since the side is decreasing... I may have failed to mention that.

OpenStudy (anonymous):

ok kool

OpenStudy (radar):

0 =(2)(9)(2) + 2sqrt (14^2 - 9^2)a \[0=36 +2a \sqrt{14^{2}-9^{2}}\]

OpenStudy (anonymous):

I got 1.67851 for the answer.

OpenStudy (radar):

@sonic, did I correctly plug in the values?

OpenStudy (anonymous):

18/sqrt(115)

OpenStudy (anonymous):

kool

OpenStudy (anonymous):

thanks for the help

OpenStudy (radar):

\[0=36 +2a \sqrt{115}\]\[2a=-36/\sqrt{115}\]\[a=-18/\sqrt{115}=\]

OpenStudy (anonymous):

@radar I don't think you put the value of db/dt as -2....you should get a positive answer.

OpenStudy (radar):

you're right I overlooked that, because it is moving down ? right?

OpenStudy (anonymous):

alright thanks i got it. hey radar can you tell me what i am doing wrong here Find the linear approximation L(x) of the function f(x)=x^8+6x^2 at a=−1. L(x)=A+Bx where A= and B= what is A? isn't A the sum of the original equation with -1 plug in plus the sum of the derivative of the equation with -1 plug in

OpenStudy (radar):

Jinnie, I am afraid that is above my pay grade lol. It looks like it can easily be differentiated, I would try it. Do you have an approved solution available? To verify?

OpenStudy (anonymous):

my software lets me know if the answer is right. thats about it

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