# Calc one related rates problem

I have encountered a related rates problem and i simply don't understand it.

a girl flying a kite holds the string 4 feet above ground level and lets out string at a rate of 2 feet per second as the kite moves horizontally at an altitude of 84 feet. find the rate at which the kite is moving horizontally with 100 feet of string has been let out

Why would there be any horizontal movement?

Thank you

• Because she is continuing to let string out. – Michael Hardy Nov 13 '16 at 16:33
• Horizontal movement is like that? <--------------------------> Releasing the the string will cause vertical movement – Gigalala Nov 13 '16 at 16:37
• Do you mean "Why would the movement be horizontal?"? That's not really part of the math problem. As the string gets longer, the kite can move to a place farther from the girl, and the problem seems to assume (unrealistically, perhaps) that the taut string is a straight line. That the direction is horizontal is given. It may also be unrealistic, but still you can find the rate. – Michael Hardy Nov 13 '16 at 16:45
• I just assumed that horizontal movement means from side to side(left to right), but i guess they mean up-down. – Gigalala Nov 13 '16 at 16:51
• They don't mean up-down, it is left-right – imranfat Nov 13 '16 at 16:52

This is an old question, but I thought I'd weigh in just in case others had a similar issue of interpretation.

Gigalala says that, upon letting out the string you would get vertical, not horizontal movement. But actually, if you think about it, you would get both. As you let out the string the kite (should) fly higher (vertical movement) and further away (horizontal movement). There are three potential sources of change over time for the sides of the right triangle, namely the change in the length of $x$ $(\frac {dx}{dt})$, the change in the length of $y$ $(\frac {dy}{dt})$, and the change in length of $z$ $(\frac {dz}{dt})$. You are given $\frac {dz}{dt}$ and asked to find $\frac {dx}{dt}$ (the horizontal component). In all likelihood the original problem statement also indicated that the vertical height $(y)$ was constant. (And, if constant, then $\frac {dy}{dt} = 0$). I'll make that assumption here, even though it may be unrealistic. (Didn't we say that the height should change as string is released? We'll just assume that, at the moment that measurements are taken, the height is fixed. Otherwise, the problem statement would have provided additional information to determine $\frac {dy}{dt}$.)

If we label the sides of a right triangle as $x$ (horizontal), $y$ (vertical), and $z$ (hypotenuse), we know that $y=80$ (remember we are drawing our triangle 4 feet off the ground) and that the rate at which the line is being released $(\frac {dz}{dt})$ is 2 ft/sec. The problem asks us to find the horizontal rate of change, $\frac {dx}{dt}$. So, differentiating with respect to time, we get

$$z^2 = x^2 + 80^2$$ $$2z \frac {dz}{dt} = 2x \frac {dx}{dt} + 0$$ $$z \frac {dz}{dt} = x \frac {dx}{dt}$$ $$\frac {z \frac {dz}{dt}}{x} = \frac {dx}{dt}$$

Substituting known values,

$$\frac {100 \cdot 2}{x} = \frac {dx}{dt}$$

The pythagorean theorem tells us that when $z=100$ and $y=80$ that $x=60$. Therefore,

$$\frac {dx}{dt} = \frac {100 \cdot 2}{60} \approx 3.33 \text{ ft/sec}$$