velocity and derivative

[wendywoo]

A 13 foot ladder is leaning against a wall so that the foot of the ladder is 1 foot from the wall. A gust of wind causes the ladder to begin sliding down the wall. The motion of the top of the ladder as it slides down the wall is described by y= -16t^2+0.05t+?168, where t is measured in seconds.

(a) when does the top of the ladder reach the ground?
(b) determine the velocity of the end of the ladder that is resting on the ground when it is 5ft from the wall. Indicate units of measure.

for part (a), i drew a diagram and since i know 2 of the triangle's sides. i was able to find the 3rd side by using the pythagorean theorem.
i know that the longest side is 13ft and the shortest side is 1foot. i solved the third side by doing 1^2+b^2=13^2 and b is 12.96ft. but i don't know what to do from here and for part (b) i just didnt understand it at all. please help!!

can you go through all the steps and explain them to me? THANKS SO MUCH!!

 

[galactus]

A 13 foot ladder is leaning against a wall so that the foot of the ladder is 1 foot from the wall. A gust of wind causes the ladder to begin sliding down the wall. The motion of the top of the ladder as it slides down the wall is described by y= -16t^2+0.05t+?168, where t is measured in seconds.

(a) when does the top of the ladder reach the ground?

It hits the ground when y=0. Set the given equation equal to 0 and solve for t.

(b) determine the velocity of the end of the ladder that is resting on the ground when it is 5ft from the wall. Indicate units of measure.

By Pythagoras, when the ladder is 5 feet from the wall, it is \(\displaystyle y=\sqrt{13^{2}-5^{2}}\) up the wall.

Set the given equation equal to this result and solve for t. Then, using this newly found t value, sub it into the velocity equation to find v(t)=dy/dt.

The velocity equation is the derivative of the given position equation.

Next, implicitly differentiate \(\displaystyle x^{2}+y^{2}=D^{2}\)

\(\displaystyle x\frac{dx}{dt}+y\frac{dy}{dt}=D\frac{dD}{dt}\)...........[1]

Note, the ladder does not change length so dD/dt=0

You have the rest of the info now. Sub it all into [1] and solve for dx/dt.

 

[wendywoo]

hmmm okay i tried to do part a and i set the y=0

so i did: -?168= -16t^2+0.05t
-12.96= -16t^2+0.05t
-12.96= 0.05(-320t^2+t)
-259.2= (-320t^2+t)

and then i don't know what to do next to solve t. help??

 

[galactus]

Use the quadratic formula.

 

[wendywoo]

can you simply explain part b?? i don't understand what the question is asking. if the ladder is laid flat on the ground where the height is 0, how come i have to use the pythagorean theorem when it's not a triangle??

 

[galactus]

can you simply explain part b?? i don't understand what the question is asking. if the ladder is laid flat on the ground where the height is 0, how come i have to use the pythagorean theorem when it's not a triangle??

part b is independent of part a. The ladder is not flat on the ground. The problem asks for how fast the bottom is slipping out when the bottom is 5 feet from the wall. Follow the outline I gave on how to solve part b and you'll be OK.

 

[wendywoo]

Thanks for the help with part a, now i finally understand it!!

for part b, you said, "By Pythagoras, when the ladder is 5 feet from the wall, it is \(\displaystyle y=\sqrt{13^{2}-5^{2}}\) up the wall.
Set the given equation equal to this result and solve for t. Then, using this newly found t value, sub it into the velocity equation to find v(t)=dy/dt."

I'm not sure I'm following. So i need to find y using the pythagorean theorem and y is 12. what do i do from here? I don't know what you mean by set this given equation to this result and solve for t. Then, use this t value and sub it into the velocity equation to find the derivative of the velocity equation?? i know I have to find the derivative of the velocity equation but what's the 12? where am i supposed to plug in the 12?!?

 

[galactus]

Thanks for the help with part a, now i finally understand it!!

for part b, you said, "By Pythagoras, when the ladder is 5 feet from the wall, it is \(\displaystyle y=\sqrt{13^{2}-5^{2}}\) up the wall.
Set the given equation equal to this result and solve for t. Then, using this newly found t value, sub it into the velocity equation to find v(t)=dy/dt."

I'm not sure I'm following. So i need to find y using the pythagorean theorem and y is 12.

Yes, it is 12.

I don't know what you mean by set this given equation to this result and solve for t.

The given equation is your position function, \(\displaystyle y=-16t^{2}+\frac{t}{20}+\sqrt{168}\)

Set this equal to 12 and solve for t:

\(\displaystyle -16t^{2}+\frac{t}{20}+\sqrt{168}-12=0\)

Use the quadratic formula.

use this t value and sub it into the velocity equation

Velocity is the derivative of the position. So, differentiate \(\displaystyle -16t^{2}+\frac{t}{20}+\sqrt{168}\)

Giving \(\displaystyle v(t)=\frac{dy}{dt}=-32t+\frac{1}{20}\). Sub the t found from before into this. This is how fast the ladder is going down the wall when the top is 12 feet from the ground.

dy/dt will be negative because the ladder is going down.

where am i supposed to plug in the 12?!?

y=12. Plug that, along with x and the dy/dt you just found, into:

\(\displaystyle x\frac{dx}{dt}+y\frac{dy}{dt}=D\frac{dD}{dt}\) and solve for dx/dt.

dx/dt is how fast the ladder is slipping away from the wall on the ground. dy/dt is the velocity or how fast it is sliding down the wall.

Remember what I said about dD/dt. That makes it easier. This stands to reason because the ladder does not change length.

 

[wendywoo]

okay. i understand everything but the last part.
How can i incorporate 12 into the implicit differentiation. what is "12"??

 

[galactus]

You appear to be hung up on that 12

The 12 is used in the end when we sub it into the differentiated Pythagorean formula (below), and near the start when the position function is set equal to 12 in order to find t.

y is 12. Remember?. 12 feet is how high up the wall to the top of the ladder. Look at the diagram I posted.

\(\displaystyle x\frac{dx}{dt}+y\frac{dy}{dt}=D\frac{dD}{dt}\)

\(\displaystyle 5\frac{dx}{dt}+12(-7.84)=13(0)\)

solve for dx/dt.

Let me know what you get.

 

[wendywoo]

The given equation is your position function, y= -16t^2+0.05t+?168.

Set this equal to 12 and solve for t: i got -16t^2+0.05t+?168-12=0
then i simplified the equation into: -16t^2+0.05t+0.96=0
and i used the quadratic formula and now i'm at = (-0.05 +/- 7.84)/-32= -0.24.
i don't understand how you got 7.84??

 

[galactus]

I don't understand how you got 7.84??

The way I have been trying to explain it....by subbing the time we find into the velocity function.

Like so:

\(\displaystyle -16t^{2}+\frac{t}{20}+\sqrt{168}=12\)

Using the quadratic formula. Use the 'minus' case. The 'plus' case is extraneous.

\(\displaystyle t=\frac{-\frac{1}{20}\pm\sqrt{(\frac{1}{20})^{2}-4(-16)(\sqrt{168}-12)}}{2(-16)}\)

\(\displaystyle t=\frac{\frac{-1}{20}-\sqrt{61.537}}{-32}=.2467\)

this works out to be .2467 seconds.

Sub this into the velocity function...the derivative of the position function.

\(\displaystyle \frac{dy}{dt}=v(t)=-32t+\frac{1}{20}\)

\(\displaystyle v(.2467)= -7.844\)

Now, sub this and the other info into \(\displaystyle x\frac{dx}{dt}+y\frac{dy}{dt}=D\frac{dD}{dt}\)

\(\displaystyle 5\frac{dx}{dt}+12(-7.844)=13(0)\)

\(\displaystyle 5\frac{dx}{dt}-94.128=0\)

\(\displaystyle \frac{dx}{dt}=\frac{94.128}{5}=18.83 \;\ ft/sec\)

The bottom is moving out at 18.83 ft/sec when the top is 5 feet off of the ground.

See now?.

 

[wendywoo]

Thank you so so much. After struggling with this problem for about two days now. I finally get it. Thank you for walking me through. It was great help.