# Changing rates

1. Mar 18, 2005

### 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.

2. Mar 18, 2005

### SpaceTiger

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

3. Mar 18, 2005

### 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.

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

4. Mar 18, 2005

### UrbanXrisis

thanks

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

that doesnt seem correct though

5. Mar 18, 2005

### 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?

6. Mar 18, 2005

### UrbanXrisis

sorry, that is in units/minutes

196.855 is the rate at which x is increasing.

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

7. Mar 18, 2005

### UrbanXrisis

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

is that what you mean my error?

8. Mar 18, 2005

### MathStudent

yes, that was it...

whats the answer you get now?

Last edited: Mar 18, 2005
9. Mar 18, 2005

### UrbanXrisis

21.333 units/min

10. Mar 18, 2005

### MathStudent

how did you come up with that answer?

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

Last edited: Mar 18, 2005
11. Mar 18, 2005

### MathStudent

by the way... did the problem tell you that y doesnt change or is that your assumption? You get a different answer if z is constant.

12. Mar 19, 2005

### UrbanXrisis

yeah, the problem tells me that y doesnt change so I have to use tan(theta)

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

Last edited: Mar 19, 2005
13. Mar 19, 2005

### MathStudent

Just to let you know, there is another way of solving this that lets 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...

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

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