# Connected rates of change kite problem

A boy is flying a kite. The boy is running into the wind at a steady speed of 3 m s−1, with the kite flying directly downwind of the boy. At one instant, the length of the string between the child and the kite is 20 m, and is increasing at a steady rate of 2 m s−1. The angle θ that the string makes with the vertical is π/4 radians, and is decreasing at a steady rate of 0.2 radians per second. Find the velocity of the kite and the angle of the velocity from the horizontal

I started the problem by considering $$\sin(\theta) = \frac{y}{20}$$ and differentiating to give $$\\cos(\theta) \frac{d\theta}{dt}= \frac{1}{20}\frac{dy}{dt}$$ which could be rearranged to give expressions for the vertical and horizontal velocities. However, these were incorrect. I'm not really sure whether I'm approaching this incorrectly. Any help would be appreciated, thanks.

We are given that the angle to the kite is $$\theta(t) = \frac \pi 4 - 0.2t$$ and that the length of the string is $$L(t) = 20 + 2t$$. If the boy starts at $$(0,0)$$ and runs to the left, we can write the following equations for the $$x$$ and $$y$$ positions of the kite: $$x(t) = -3t + L(t)*\cos(\theta(t))$$ $$y(t) = L(t)*\sin(\theta(t))$$ Now you just need to differentiate.
• Thanks, that works nicely. I think the question was a bit ambiguous regarding where $\theta$ was measured from initially, so the expressions for $y(t)$ and $x(t)$ should have the $\cos(\theta)$ and $\sin(\theta)$ swapped. – bebop Dec 21 '18 at 17:54
• You're welcome. And you're right, I overlooked that $\theta$ was measured according to the vertical. If you accept the answer, please give it a checkmark so it won't remain an unanswered question. – Jens Dec 21 '18 at 18:06
If $$t$$ it the time in seconds you are given that $$\theta = \frac \pi 4 - 0.2t$$. If we measure $$x$$ horizontally from underneath where the kite starts, the boy starts at $$x=20 \cos \left(\frac \pi 4\right)$$, so his position is $$x(t)=20 \cos \left(\frac \pi 4\right)+3t$$. Can you figure out the position of the kite as a function of $$t$$?
• Thanks for the response. So if you were to construct a triangle where $\tan(\theta-0.2t) = \frac{y(t)}{x(t)}$ where $x(t)=20 \cos \left(\frac \pi 4\right)+3t$ you could gain an expression for the position with respect to time, which could be differentiated to find an expression for the velocity. This doesn't seem to give the correct answer. Am I missing something? – bebop Dec 21 '18 at 16:45
• The kite does not stay at $x=0$. It is moving, too. You need to compute that based on the angle and the length of the string. – Ross Millikan Dec 21 '18 at 17:00