Projectile problem

[trainee engineer]

Good Afternoon everyone

I am learning calculus for the first time and have come across a problem in my homework assignment that is giving me some trouble.
The problem is as follows:

A projectile with co-ordinates (x, y) is moving along a parabolic trajectory described by the equation [math]2(y+2)=(x+2)^2[/math]. At what point on the trajectory is the height (y) changing at the same rate as the distance (x) from the projectile's point of origin.

I have been stuck with this but I first rearranged the equation for both x and y

[math]y=(x^2)/2+2x[/math] and [math]x=\sqrt{2(y+2)}-2[/math]
Then I derived both to get the equations for the rate of change for each variable with respect to the other.

[math]\frac{dy}{dx}=x+2[/math] and [math]\frac{dx}{dy}=\frac{1}{\sqrt(2(y+2))}[/math]
I was able to create a quadratic to use to solve for y and then substituted back to get x.

This has given me several possible solutions but I am confused and don't understand this.

I have attached my working in the files.

 

[topsquark]

Good Afternoon everyone

I am learning calculus for the first time and have come across a problem in my homework assignment that is giving me some trouble.
The problem is as follows:

A projectile with co-ordinates (x, y) is moving along a parabolic trajectory described by the equation [math]2(y+2)=(x+2)^2[/math]. At what point on the trajectory is the height (y) changing at the same rate as the distance (x) from the projectile's point of origin.

I have been stuck with this but I first rearranged the equation for both x and y

[math]y=(x^2)/2+2x[/math] and [math]x=\sqrt{2(y+2)}-2[/math]
Then I derived both to get the equations for the rate of change for each variable with respect to the other.

[math]\frac{dy}{dx}=x+2[/math] and [math]\frac{dx}{dy}=\frac{1}{\sqrt(2(y+2))}[/math]
I was able to create a quadratic to use to solve for y and then substituted back to get x.

This has given me several possible solutions but I am confused and don't understand this.

I have attached my working in the files.

Your whole approach is, unfortunately, not going to work. You are taking the derivative with respect to the wrong thing.

Projectile motion is motion in time. So the condition is that the rate of change of y with respect to time is the same as the rate of change of x with respect to time. ie. [imath]\dfrac{dy}{dt} = \dfrac{dx}{dt}[/imath].

So to start you off: Take the time derivative of both sides of 2(y + 2) = (x + 2)^2 and set the derivatives equal to each other. Let us know how it goes.

-Dan

 

[trainee engineer]

Your whole approach is, unfortunately, not going to work. You are taking the derivative with respect to the wrong thing.

Projectile motion is motion in time. So the condition is that the rate of change of y with respect to time is the same as the rate of change of x with respect to time. ie. [imath]\dfrac{dy}{dt} = \dfrac{dx}{dt}[/imath].

So to start you off: Take the time derivative of both sides of 2(y + 2) = (x + 2)^2 and set the derivatives equal to each other. Let us know how it goes.

-Dan

So I do the following?

[math]2(y+2)=(x+2)^2[/math]
[math]\frac{d(2y+4)}{dt}=\frac{d(x+2)^2}{dt}[/math]
[math]2\frac{dy}{dt}=(2x+4)\frac{dx}{dt}[/math]
and as [math]\frac{dy}{dt}=\frac{dx}{dt}[/math]
then [math]2 = (2x+4)[/math]

 

[Deleted member 4993]

So I do the following?

[math]2(y+2)=(x+2)^2[/math]
[math]\frac{d(2y+4)}{dt}=\frac{d(x+2)^2}{dt}[/math]
[math]2\frac{dy}{dt}=(2x+4)\frac{dx}{dt}[/math]
and as [math]\frac{dy}{dt}=\frac{dx}{dt}[/math]
then [math]2 = (2x+4)[/math]

You calculated the value of 'x' (when dx/dt = dy/dt). Now, from that calculate the value of 'y'.

 

[trainee engineer]

You calculated the value of 'x' (when dx/dt = dy/dt). Now, from that calculate the value of 'y'.

Thank you

 

[topsquark]

So I do the following?

[math]2(y+2)=(x+2)^2[/math]
[math]\frac{d(2y+4)}{dt}=\frac{d(x+2)^2}{dt}[/math]
[math]2\frac{dy}{dt}=(2x+4)\frac{dx}{dt}[/math]
and as [math]\frac{dy}{dt}=\frac{dx}{dt}[/math]
then [math]2 = (2x+4)[/math]

As Subhotosh Khan implies.. So far so good!

-Dan