Inverse Trigonometric Functions / Rates of change with 2 points of references

In summary: In this case, I think the formula is x^2 + y^2 = 4000000 (by the Pythagorean theorem). Differentiating implicitly, 2x*dx/dt + 2y*dy/dt= 0. You know dx/dt (the rate of change of x) and you want to find dy/dt.
  • #1
kioria
54
0
Hi,

A questions regarding trigonometric (or inverse) functions. I just can't get it right.

Q1: A picture 2 m high is hung on the wall with its bottom 6 m above the observer's eye level. How far should the viewer stand for the picture to subtend the largest possible vertical angle with the viewer's eye?

Thanks

Also, I would like to know more about (in basic level, I am a first year) the rates of change. I cannot give a specific topic of what I am thinking, but I can give an example:

E1: A light house (P) is located at the middle of a long straight beach with its light-beam rotating at 3 rpm. And 2000 m directly below the location of P, a ship (S) travels towards the P at 20 km/s. How fast is the light moving at point S at this moment?

(Please don't try to do this, as I made this thing up).

If you know the topic, please reply and just leave the topic name (and by any chance you know a great site explaining it, then please tell me the site as well!)

Thanks :)
 
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  • #2
kioria said:
Hi,

A questions regarding trigonometric (or inverse) functions. I just can't get it right.

Q1: A picture 2 m high is hung on the wall with its bottom 6 m above the observer's eye level. How far should the viewer stand for the picture to subtend the largest possible vertical angle with the viewer's eye?

Draw a picture. Draw the picture itself, the lines from the top and bottom of the picture to the viewer's eye, and the line from the from the viewer's eye perpendicular to the wall. You now have two right triangles. The base of both triangles has length x, the unknown distance from the viewer to the wall.
The height of the larger triangle is 8 m (that's one heck of a high wall!) so tan(θ)= 8/x (θ is the larger angle at the viewer's eye).
The height of the smaller triangle is 6 m so tan(φ)= 6/x (φ is the smaller angle at the viewer's eye.)

The angle subtended is θ-&phi= tan-1(8/x)- tan-1(6/x).

Can you find x to maximize that?


kioria said:
Also, I would like to know more about (in basic level, I am a first year) the rates of change. I cannot give a specific topic of what I am thinking, but I can give an example:

E1: A light house (P) is located at the middle of a long straight beach with its light-beam rotating at 3 rpm. And 2000 m directly below the location of P, a ship (S) travels towards the P at 20 km/s. How fast is the light moving at point S at this moment?

(Please don't try to do this, as I made this thing up).

If you know the topic, please reply and just leave the topic name (and by any chance you know a great site explaining it, then please tell me the site as well!)

Thanks :)

It's not clear what you want to know. Typically this is referred to as "related rates" and is based on the chain rule. Set up a "static" formula relating the two variables and differentiate with respect to time to get a formula relating the rates of change of the two variables.
 
  • #3



Hello,

Inverse trigonometric functions can be tricky, but with practice and understanding, they can be easily mastered. In this question, we are given the height of the picture (2 m) and the distance of the bottom of the picture from the observer's eye level (6 m). To find the distance the viewer should stand for the largest possible vertical angle, we need to use the inverse trigonometric function, specifically the arctan function. The formula for this is arctan (opposite/adjacent). In this case, the opposite side is the height of the picture (2 m) and the adjacent side is the distance from the bottom of the picture to the observer's eye level (6 m). So, the formula would be arctan (2/6). This gives us an angle of approximately 18.4 degrees. This means that the viewer should stand at a distance where the angle between the viewer's eye and the top of the picture is 18.4 degrees for the picture to subtend the largest possible vertical angle.

As for rates of change, it is a concept that deals with how one variable changes in relation to another variable. In your example, the rate of change is the speed at which the light beam is moving at a specific point (S) when the ship is moving towards the lighthouse at a constant speed. This can be calculated using the derivative, which is a mathematical tool used to find rates of change. In this case, the derivative of the angular velocity (3 rpm) would give us the rate of change of the light beam's speed at point S.

I cannot give you a specific topic for rates of change, as it is a broad concept used in many fields of mathematics. However, some common applications include finding the speed or acceleration of an object, calculating the slope of a curve, and determining the rate of growth or decay of a population. As for a great site explaining rates of change, I would recommend Khan Academy or MathIsFun. They have clear and concise explanations with examples for various topics in mathematics.

I hope this helps and good luck with your studies!
 

1. What are inverse trigonometric functions?

Inverse trigonometric functions are mathematical operations that can be used to find the angle at which a specific trigonometric ratio occurs. For example, the inverse sine function can be used to find the angle whose sine is equal to a given value.

2. How can inverse trigonometric functions be used in real life?

Inverse trigonometric functions are commonly used in navigation and surveying to calculate angles and distances. They are also used in physics and engineering to solve problems involving triangles and waves.

3. What is the relationship between inverse trigonometric functions and regular trigonometric functions?

The inverse trigonometric functions are the inverse operations of the regular trigonometric functions. This means that they "undo" the original trigonometric function and return the angle that produced a given value.

4. How do you find the derivative of an inverse trigonometric function?

To find the derivative of an inverse trigonometric function, you can use the chain rule. For example, to find the derivative of inverse sine, you would take the derivative of the inside function (in this case, the sine function) and multiply it by the derivative of the inverse function (in this case, 1/(√(1-x^2))).

5. How do you find the rate of change between two points using inverse trigonometric functions?

To find the rate of change between two points using inverse trigonometric functions, you would first find the angle at each point using the inverse function. Then, you can use the slope formula (change in y / change in x) to find the rate of change between the two points.

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