Find the distance traveled of a piece of prey dropped by a hawk.

  • #1

Homework Statement

So the question states that a hawk is flying at an altitude of 180m at 15m/s and drops its prey. The parabolic motion of the prey is defined by the function 180-x^2/45 were x is the horizontal distance traveled in meters. Solve for the total distance traveled from the moment it was dropped to when it hits the ground.



Homework Equations


y= 180 - x^2/45
L(arclength)= ∫[itex]\sqrt{1+ (dy/dx)^2}[/itex]dx



The Attempt at a Solution


I began by setting the bounds at 180 and 0, since this is the interval of height were looking at, and switched my function to solve for x. So x= -(y^1/2)/2. I then solved for dx/dy and found it to be -1/4y^1/2.

I then used the other form of the arclength where I set x=g(y) so I could use dx/dy and the bounds I had selected. I get the integral set at ∫[itex]\sqrt{1+(1/16y^2)}[/itex] dy, and I don't know where to go from there.

I saw that most of the time you want it so that you can either use a u substitution or factor the quantity under the square root so that you can take the square root of a square and then have a simple integral, however I'm not sure as to how to go about doing that.

Sorry for the format, I'm not good at using the different functions of the website. If someone could give me a pointer or let me know if I've gone wrong at some point in the calculation I'd very much appreciate it.
 
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Answers and Replies

  • #2
SammyS
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Homework Statement

So the question states that a hawk is flying at an altitude of 180m at 15m/s and drops its prey. The parabolic motion of the prey is defined by the function 180-x^2/45 were x is the horizontal distance traveled in meters. Solve for the total distance traveled from the moment it was dropped to when it hits the ground.

Homework Equations


y= 180 - x^2/45
L(arclength)= ∫[itex]\sqrt{1+ (dy/dx)^2}[/itex]dx

The Attempt at a Solution


I began by setting the bounds at 180 and 0, since this is the interval of height were looking at, and switched my function to solve for x. So x= -(y^1/2)/2. I then solved for dx/dy and found it to be -1/4y^1/2.

I then used the other form of the arclength where I set x=g(y) so I could use dx/dy and the bounds I had selected. I get the integral set at ∫[itex]\sqrt{1+(1/16y^2)}[/itex] dy, and I don't know where to go from there.

I saw that most of the time you want it so that you can either use a u substitution or factor the quantity under the square root so that you can take the square root of a square and then have a simple integral, however I'm not sure as to how to go about doing that.

Sorry for the format, I'm not good at using the different functions of the website. If someone could give me a pointer or let me know if I've gone wrong at some point in the calculation I'd very much appreciate it.
While you do need to solve y= 180 - x^2/45 for x to get the limits of integration, there is no need to find dx/dy .

You integral has dy/dx in it, not dx/dy .
 

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