Rate of Change of x in a Right Triangle

[UrbanXrisis]

In a right triangle with sides x,y,z, the theta between leg z and leg y is increasing at a constant rate of 3 rad/min. What is the rate at which x is increasing in units per minute when x equals 3 units and z is 5 units.

so the triangle is basically a 3,4,5 triangle. The theta is between the leg 5 and 4. Leg 4 does not chage since it's the base. To set up a changing rates problem, I did:

tan\theta = \frac{O}{A}
4tan \frac{d \theta}{dt} = \frac{dx}{dt}
\theta=-0.570 rad/min

I think there is something wrong with this but I'm not sure why the theta is negative.

 

[SpaceTiger]

tan\theta = \frac{O}{A}
4tan \frac{d \theta}{dt} = \frac{dx}{dt}

Are you sure the derivative of tan(\theta) is tan (d \theta/dt)?

 

[MathStudent]

from your post it seems that you are dealing with a right triangle where z is the hypotenuse, y is the adjacent, and x is the opposite, with z=5, y=4, x=3.

your derivative is incorrect.

tan is a function of theta, and theta is a function of time, so tan is actually a composite function of time.

You have to use the chairule when differentiating tan wrt t ie:

if f(\theta) = tan \theta then

\frac{df}{dt} = \frac{df}{d \theta} \frac{d \theta}{dt}

just an added note: you can assume that either y or z is constant, and they both work out the same.

edit: spacetiger beat me to it

 

[UrbanXrisis]

thanks

tan\theta = \frac{O}{A}
4sec^2 \frac{d \theta}{dt} = \frac{dx}{dt}
\theta=196.855 rad/min

that doesn't seem correct though

 

[MathStudent]

your solving for the wrong quantity... the question asks for the rate at which x increases. The rate of change of theta is already given.

also your derivative is still incorrect,, there is a small error, do you see it?

 

[UrbanXrisis]

sorry, that is in units/minutes

196.855 is the rate at which x is increasing.

tan\theta = \frac{O}{A}
4sec^2 \frac{d \theta}{dt} = \frac{dx}{dt}
4 \frac{1}{tan(3)^2}= \frac{dx}{dt}
\frac{dx}{dt}=196.855 units/min

 

[UrbanXrisis]

tan\theta = \frac{O}{A}
4sec^2 \theta \frac{d \theta}{dt} = \frac{dx}{dt}

is that what you mean my error?

 

[MathStudent]

yes, that was it...

whats the answer you get now?

 

[UrbanXrisis]

21.333 units/min

 

[MathStudent]

how did you come up with that answer?

hint: your solving for \frac{dx}{dt} [/itex], so you need both the values of \frac{d \theta}{dt} and sec\theta

 

[MathStudent]

by the way... did the problem tell you that y doesn't change or is that your assumption? You get a different answer if z is constant.

 

[UrbanXrisis]

yeah, the problem tells me that y doesn't change so I have to use tan(theta)

tan\theta=\frac{O}{A}
4sec^2\theta \frac{d\theta}{dt}=\frac{dx}{dt}
\theta=sin^{-1}\frac{3}{5}=36.87
\frac{4}{cos^236.87} 3rad/min =\frac{dx}{dt}
\frac{dx}{dt}=18.750units/min

sorry, I made a mistake, this is my final answer

 

[MathStudent]

Thats the right answer.

Just to let you know, there is another way of solving this that let's you avoid having to find what theta is, by using the definitions of sec(theta) for a right triangle.

cos \theta = \frac{A}{H}

sec \theta = \frac{1}{cos \thet} = \frac{H}{A}

\Rightarrow sec^2 \theta = \frac{H^2}{A^2}

Therefore the problem boils down to simple arithmetic...

\frac{dx}{dt} = 4sec^2\theta \frac{d\theta}{dt}

\Rightarrow \frac{dx}{dt} = 4 (\frac{5^2}{4^2})(3)