MHB What is the angle and rate of change when flying a kite?

[Petrus]

You fly a kite. The dragon is 15m away from you (measured horizontally) and
moves right now straight up. The rope is rolled out with the speed of 4 dm / s, and 30m
is unwound. (a) Draw a clear picture of the situation, and explain what all introduced
designations stand for.
(b) What is the angle between the line and ground at this moment?
(c) How fast the angle changes? Also calculate the value to degrees per second
and consider whether it is reasonable.my progress:
(a)

(b)$$\cos^{-2}( \frac{15}{30})=60degree$$
(c) i need help with this part I know that dz/dt=0.4m/s

Regards,
$$|\pi\rangle$$

 

[I like Serena]

(c) i need help with this part I know that dz/dt=0.4m/s

Hey Petrus! :)

Perhaps you can draw a second picture that shows the situation some time $t$ later?

And then calculate the angle in this situation?

Btw, your assumption that $dz/dt=0.4 \text{ m/s}$ is incorrect.
The problem statement says that the line gets unwound at $0.4 \text{ m/s}$.
That is not the same as the height increasing at this rate.

 

[Petrus]

Hey Petrus! :)

Perhaps you can draw a second picture that shows the situation some time $t$ later?

And then calculate the angle in this situation?

Btw, your assumption that $dz/dt=0.4 \text{ m/s}$ is incorrect.
The problem statement says that the line gets unwound at $0.4 \text{ m/s}$.
That is not the same as the height increasing at this rate.

do you mean like 2 second later the hypotenuse is 30.8m and the bottom 15.8m did i get it correct?

Regards,
$$|\pi\rangle$$

 

[I like Serena]

do you mean like 2 second later the hypotenuse is 30.8m and the bottom 15.8m did i get it correct?

The hypotenuse would indeed become 30.8 m.
But according to your problem statement the kite rises straight up meaning the bottom would still be 15 m.

Btw, I would try to generalize it for an arbitrary time $t$.
If you have the angle as function of $t$ you can then differentiate it.

 

[Petrus]

The hypotenuse would indeed become 30.8 m.
But according to your problem statement the kite rises straight up meaning the bottom would still be 15 m.

Btw, I would try to generalize it for an arbitrary time $t$.
If you have the angle as function of $t$ you can then differentiate it.

do you mean like this $$cos(x)=\frac{30+0.4t}{15}$$ so i got $$x=\cos^{-1}(\frac{30+0.4t}{15})$$

Regards,
$$|\pi\rangle$$

 

[I like Serena]

do you mean like this $$cos(x)=\frac{30+0.4t}{15}$$ so i got $$x=\cos^{-1}(\frac{30+0.4t}{15})$$

Regards,
$$|\pi\rangle$$

Almost! ;)

It should be:
$$x=\cos^{-1}\left(\frac{15}{30+0.4t}\right)$$

Now you can take the derivative to find the rate of change of the angle.

 

[Petrus]

Almost! ;)

It should be:
$$x=\cos^{-1}\left(\frac{15}{30+0.4t}\right)$$

Now you can take the derivative to find the rate of change of the angle.

Yep i mean that. So if we derivate it we get this 37.5'/''(''('75.'+'1. t')''^'2 sqrt'('1-1406.25'/''('75.'+'1. t')''^'2')'')' - Wolfram|Alpha
what do you mean after?

Regards,
$$|\pi\rangle$$

 

[I like Serena]

Yep i mean that. So if we derivate it we get this 37.5'/''(''('75.'+'1. t')''^'2 sqrt'('1-1406.25'/''('75.'+'1. t')''^'2')'')' - Wolfram|Alpha
what do you mean after?

Huh?
Which "after"?

Anyway, you can substitute $t=0$ to find the current rate of change of the angle.

See Wolfram|Alpha here.

 

[Petrus]

Huh?
Which "after"?

Anyway, you can substitute $t=0$ to find the current rate of change of the angle.

See Wolfram|Alpha here.

yes i get that, but i have hard understanding what is the unit. I mean this is not degree is it m/s? or i am lost..
edit: if i get it correct then after 0 secund its that much in meter?
Regards,
$$|\pi\rangle$$

 

[I like Serena]

yes i get that, but i have hard understanding what is the unit. I mean this is not degree is it m/s? or i am lost..
edit: if i get it correct then after 0 secund its that much in meter?

We're talking about the rate of change of the angle.
By default W|A will use the unit $\text{rad/s}$.

 

[Petrus]

We're talking about the rate of change of the angle.
By default W|A will use the unit $\text{rad/s}$.

ok so we got 0.44106 degree/sec right?

Regards,
$$|\pi\rangle$$

 

[I like Serena]

ok so we got 0.44106 degree/sec right?

Yep! :D