Speed of car pulling a rope
A car is being pulled along by a rope attached to the tow-bar at the back of the car.
The rope passes through a pulley, the top of which is $3$ metres further from the ground than the tow-bar.
The pulley is $x$ metres horizontally from the tow-bar.
The rope is being winched in at a speed of $0.6 ; text{ms}^{-1}$.
The wheels of the car remain in contact with the ground.
At what speed is the car moving when the length of the rope, $L$, between the tow-bar and the pulley is $5.4$ metres?
I drew a right-angle triangle with the hypotenuse being $L$, adjacent being $x$ and tangent being $3$.
We know that the speed of the rope is $0.6$, i.e. $frac{dR}{dt} = 0.6$.
Not too sure what to do next by it seems the chain rule is gonna be used somewhere.
Thanks!!!
derivatives
asked Nov 25 at 0:51
squenshl
11
add a comment |
A car is being pulled along by a rope attached to the tow-bar at the back of the car.
The rope passes through a pulley, the top of which is $3$ metres further from the ground than the tow-bar.
The pulley is $x$ metres horizontally from the tow-bar.
The rope is being winched in at a speed of $0.6 ; text{ms}^{-1}$.
The wheels of the car remain in contact with the ground.
At what speed is the car moving when the length of the rope, $L$, between the tow-bar and the pulley is $5.4$ metres?
I drew a right-angle triangle with the hypotenuse being $L$, adjacent being $x$ and tangent being $3$.
We know that the speed of the rope is $0.6$, i.e. $frac{dR}{dt} = 0.6$.
Not too sure what to do next by it seems the chain rule is gonna be used somewhere.
Thanks!!!
derivatives
asked Nov 25 at 0:51
squenshl
11
You have a reasonable amount of explanatory text, here, but a picture would really set up good understanding.
– Adrian Keister
Nov 25 at 1:05
$frac{dR}{dt} = -0.6$ I believe. The length of the rope is decreasing.
– saulspatz
Nov 25 at 1:38
What you have said is correct, except for the sign error I've already pointed out. You need to solve for $x$ in terms of $R$, which is easy, given your triangle, and then apply the chain rule. Try that, and come back if it doesn't work out for you.
– saulspatz
Nov 25 at 1:41
Using Pythagoras' Theorem $x = sqrt{L^2-9}$ so $frac{dx}{dL} = frac{L}{sqrt{L^2-9}}$ so using the chain rule $$frac{dx}{dt} = frac{dx}{dL}times frac{dL}{dt} = frac{-0.6L}{sqrt{L^2-9}}.$$ Throw in $L = 5.4$ and we are done.
– squenshl
Nov 25 at 4:25
add a comment |
A car is being pulled along by a rope attached to the tow-bar at the back of the car.
The rope passes through a pulley, the top of which is $3$ metres further from the ground than the tow-bar.
The pulley is $x$ metres horizontally from the tow-bar.
The rope is being winched in at a speed of $0.6 ; text{ms}^{-1}$.
The wheels of the car remain in contact with the ground.
At what speed is the car moving when the length of the rope, $L$, between the tow-bar and the pulley is $5.4$ metres?
I drew a right-angle triangle with the hypotenuse being $L$, adjacent being $x$ and tangent being $3$.
We know that the speed of the rope is $0.6$, i.e. $frac{dR}{dt} = 0.6$.
Not too sure what to do next by it seems the chain rule is gonna be used somewhere.
Thanks!!!
derivatives
asked Nov 25 at 0:51
squenshl
11
A car is being pulled along by a rope attached to the tow-bar at the back of the car.
The rope passes through a pulley, the top of which is $3$ metres further from the ground than the tow-bar.
The pulley is $x$ metres horizontally from the tow-bar.
The rope is being winched in at a speed of $0.6 ; text{ms}^{-1}$.
The wheels of the car remain in contact with the ground.
At what speed is the car moving when the length of the rope, $L$, between the tow-bar and the pulley is $5.4$ metres?
I drew a right-angle triangle with the hypotenuse being $L$, adjacent being $x$ and tangent being $3$.
We know that the speed of the rope is $0.6$, i.e. $frac{dR}{dt} = 0.6$.
Not too sure what to do next by it seems the chain rule is gonna be used somewhere.
Thanks!!!
derivatives
derivatives
asked Nov 25 at 0:51
squenshl
11
asked Nov 25 at 0:51
squenshl
11
asked Nov 25 at 0:51
squenshl
11
asked Nov 25 at 0:51
squenshl
11
asked Nov 25 at 0:51
squenshl
11
11
You have a reasonable amount of explanatory text, here, but a picture would really set up good understanding.
– Adrian Keister
Nov 25 at 1:05
$frac{dR}{dt} = -0.6$ I believe. The length of the rope is decreasing.
– saulspatz
Nov 25 at 1:38
What you have said is correct, except for the sign error I've already pointed out. You need to solve for $x$ in terms of $R$, which is easy, given your triangle, and then apply the chain rule. Try that, and come back if it doesn't work out for you.
– saulspatz
Nov 25 at 1:41
Using Pythagoras' Theorem $x = sqrt{L^2-9}$ so $frac{dx}{dL} = frac{L}{sqrt{L^2-9}}$ so using the chain rule $$frac{dx}{dt} = frac{dx}{dL}times frac{dL}{dt} = frac{-0.6L}{sqrt{L^2-9}}.$$ Throw in $L = 5.4$ and we are done.
– squenshl
Nov 25 at 4:25
add a comment |
You have a reasonable amount of explanatory text, here, but a picture would really set up good understanding.
– Adrian Keister
Nov 25 at 1:05
$frac{dR}{dt} = -0.6$ I believe. The length of the rope is decreasing.
– saulspatz
Nov 25 at 1:38
What you have said is correct, except for the sign error I've already pointed out. You need to solve for $x$ in terms of $R$, which is easy, given your triangle, and then apply the chain rule. Try that, and come back if it doesn't work out for you.
– saulspatz
Nov 25 at 1:41
Using Pythagoras' Theorem $x = sqrt{L^2-9}$ so $frac{dx}{dL} = frac{L}{sqrt{L^2-9}}$ so using the chain rule $$frac{dx}{dt} = frac{dx}{dL}times frac{dL}{dt} = frac{-0.6L}{sqrt{L^2-9}}.$$ Throw in $L = 5.4$ and we are done.
– squenshl
Nov 25 at 4:25
You have a reasonable amount of explanatory text, here, but a picture would really set up good understanding.
– Adrian Keister
Nov 25 at 1:05
You have a reasonable amount of explanatory text, here, but a picture would really set up good understanding.
– Adrian Keister
Nov 25 at 1:05
$frac{dR}{dt} = -0.6$ I believe. The length of the rope is decreasing.
– saulspatz
Nov 25 at 1:38
$frac{dR}{dt} = -0.6$ I believe. The length of the rope is decreasing.
– saulspatz
Nov 25 at 1:38
What you have said is correct, except for the sign error I've already pointed out. You need to solve for $x$ in terms of $R$, which is easy, given your triangle, and then apply the chain rule. Try that, and come back if it doesn't work out for you.
– saulspatz
Nov 25 at 1:41
What you have said is correct, except for the sign error I've already pointed out. You need to solve for $x$ in terms of $R$, which is easy, given your triangle, and then apply the chain rule. Try that, and come back if it doesn't work out for you.
– saulspatz
Nov 25 at 1:41
Using Pythagoras' Theorem $x = sqrt{L^2-9}$ so $frac{dx}{dL} = frac{L}{sqrt{L^2-9}}$ so using the chain rule $$frac{dx}{dt} = frac{dx}{dL}times frac{dL}{dt} = frac{-0.6L}{sqrt{L^2-9}}.$$ Throw in $L = 5.4$ and we are done.
– squenshl
Nov 25 at 4:25
Using Pythagoras' Theorem $x = sqrt{L^2-9}$ so $frac{dx}{dL} = frac{L}{sqrt{L^2-9}}$ so using the chain rule $$frac{dx}{dt} = frac{dx}{dL}times frac{dL}{dt} = frac{-0.6L}{sqrt{L^2-9}}.$$ Throw in $L = 5.4$ and we are done.
– squenshl
Nov 25 at 4:25
add a comment |
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You have a reasonable amount of explanatory text, here, but a picture would really set up good understanding.
– Adrian Keister
Nov 25 at 1:05
$frac{dR}{dt} = -0.6$ I believe. The length of the rope is decreasing.
– saulspatz
Nov 25 at 1:38
What you have said is correct, except for the sign error I've already pointed out. You need to solve for $x$ in terms of $R$, which is easy, given your triangle, and then apply the chain rule. Try that, and come back if it doesn't work out for you.
– saulspatz
Nov 25 at 1:41
Using Pythagoras' Theorem $x = sqrt{L^2-9}$ so $frac{dx}{dL} = frac{L}{sqrt{L^2-9}}$ so using the chain rule $$frac{dx}{dt} = frac{dx}{dL}times frac{dL}{dt} = frac{-0.6L}{sqrt{L^2-9}}.$$ Throw in $L = 5.4$ and we are done.
– squenshl
Nov 25 at 4:25