rates of change: A ladder 10 ft long leans against a vertica

[tessec]

I am in Calc 2 at the moment, and I am having trouble recalling a lot of concepts from Calc I.

Problem: A ladder 10 ft long leans against a vertical wall. If the bottom of the ladder slides away at a speed of 2 ft/s, how fast is the angle between the ladder and the wall changing when the bottom of the ladder is 6 ft from the base of the wall?

This problem is under the unit: inverse functions, by the way. I know that the anwer is 1/4 rad/s, I just can't figure out how to get there.

Work I have so far: not much.

I let x be the distance of the bottom of the ladder to the wall, and dx be the speed that the distance is changing. I let y be the distance from the top of the ladder to the wall, and likewise dy is the speed that the y is changing. z is the angle; dz, its rate of change.

so i want dz when x=6, dx=2, and y=8.

I know that I have to find a relation between the 3 variables, so I put tan y=x/z, but I don't know if thats correct.

help!

 

[daon]

Re: rates of change

I am in Calc 2 at the moment, and I am having trouble recalling a lot of concepts from Calc I.

Problem: A ladder 10 ft long leans against a vertical wall. If the bottom of the ladder slides away at a speed of 2 ft/s, how fast is the angle between the ladder and the wall changing when the bottom of the ladder is 6 ft from the base of the wall?

This problem is under the unit: inverse functions, by the way. I know that the anwer is 1/4 rad/s, I just can't figure out how to get there.

Work I have so far: not much.

I let x be the distance of the bottom of the ladder to the wall, and dx be the speed that the distance is changing. I let y be the distance from the top of the ladder to the wall, and likewise dy is the speed that the y is changing. z is the angle; dz, its rate of change.

so i want dz when x=6, dx=2, and z=8.

I know that I have to find a relation between the 3 variables, so I put tan y=x/z, but I don't know if thats correct.

help!

Using tan(z) = x/y you can get the angle at a specific instant (but it is already given). With the three variavles you were given, use the fact that 10^2 = x^2 + y^2 to get your y for when x is 6 (for this problem, this is not necessary).

What you need to do is get f(z)=f(x) to differentiate. To do this, use your pathagorean equation above to get 'y' by itself and substitute this into the other. Once you only have z and x in an equation, you need to get dz/dt. Notice that dz/dt = dz/dx * dx/dt.

-Daon

 

[galactus]

You want \(\displaystyle \L\\\frac{d{\theta}}{dt}\) when \(\displaystyle \L\\\frac{dx}{dt}=2\)

\(\displaystyle \L\\sin({\theta})=\frac{x}{10}\)

\(\displaystyle \L\\cos({\theta})\frac{d{\theta}}{dt}=\frac{1}{10}\frac{dx}{dt}\)....[1]

\(\displaystyle \L\\y=\sqrt{10^{2}-6^{2}}=8\)

Therefore, \(\displaystyle \L\\cos({\theta})=\frac{8}{10}=\frac{4}{5}\)

Solve [1] for \(\displaystyle \frac{d{\theta}}{dt}\)

\(\displaystyle \L\\\frac{d{\theta}}{dt}=\frac{\frac{1}{10}\frac{dx}{dt}}{\frac{4}{5}}=\frac{1}{4}\;\ rad/sec\)

 

[skeeter]

Problem: A ladder 10 ft long leans against a vertical wall. If the bottom of the ladder slides away at a speed of 2 ft/s, how fast is the angle between the ladder and the wall changing when the bottom of the ladder is 6 ft from the base of the wall?

This problem is under the unit: inverse functions, by the way. I know that the anwer is 1/4 rad/s, I just can't figure out how to get there.

\(\displaystyle \L \theta = arcsin(\frac{x}{10})\)

\(\displaystyle \L \frac{d\theta}{dt} = \frac{1}{sqrt{100 - x^2}} \frac{dx}{dt} = (\frac{1}{8})(2) = \frac{1}{4}\) rad/sec