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Calculus question - involving trigonometry
- Thread starter rshen5
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In summary, the conversation is about a physics problem involving finding the intensity of light at a certain point. The problem involves using the Pythagorean theorem and trigonometric identities to arrive at the equation for finding the maximum intensity. The person asking for help was initially confused but eventually understood the solution.
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tiny-tim
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welcome to pf!
hi rshen5! welcome to pf!
(have a theta: θ and a square-root: √)
show us what you've tried, and where you're stuck, and then we'll know how to help …
hi rshen5! welcome to pf!
(have a theta: θ and a square-root: √
)show us what you've tried, and where you're stuck, and then we'll know how to help …
start with part a
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rshen5
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you have any idea where the 8 came from ?
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tiny-tim
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rshen5 said:
i expect it's written in tiny print on the light bulb
…get on with it!
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tiny-tim
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(you don't need to know what I is, you only need to know when it's a maximum! )
why are you bothering with Pythagoras?
you've found that AN = xsinθ, and you know that AN = 1
)why are you bothering with Pythagoras?
you've found that AN = xsinθ, and you know that AN = 1
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HallsofIvy
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The problem tells you that "I" is the "illumination"- the intensity of the light at that point. However, as tinytim suggests, knowing the specific physical meaning of variables is not necessary to do the mathematics!
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tiny-tim
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HallsofIvy said:
maths is so cool!
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lurflurf
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recall from elementary trigonometry that
1+cot(t)^2=csc(t)^2
you got a little carried away
you wrote AN=x sin(theta)
also AN=1 so x=csc(theta)=1/sin(theta)
you can write I in terms of theta by substituting x=csc(theta) into I
x^2=x^2cos(theta)^2+x^2sin(theta)^2
reduces to the un helpful
x^2=x^2
not x^2=2x^2+1
1+cot(t)^2=csc(t)^2
you got a little carried away
you wrote AN=x sin(theta)
also AN=1 so x=csc(theta)=1/sin(theta)
you can write I in terms of theta by substituting x=csc(theta) into I
x^2=x^2cos(theta)^2+x^2sin(theta)^2
reduces to the un helpful
x^2=x^2
not x^2=2x^2+1
- #10
FAQ: Calculus question - involving trigonometry
What is the relationship between calculus and trigonometry?
Calculus and trigonometry are closely related fields of mathematics. Trigonometry deals with the study of triangles and their properties, while calculus deals with the study of functions and their rates of change. Trigonometric functions are used extensively in calculus to model and solve real-world problems.
How are trigonometric functions used in calculus?
Trigonometric functions such as sine, cosine, and tangent are used in calculus to model and solve problems involving rates of change, such as velocity and acceleration. They are also used to find the area under curves and to calculate integrals.
What are some common applications of calculus with trigonometry?
Calculus with trigonometry is used in a variety of fields, including physics, engineering, and economics. It is used to model and analyze various phenomena, such as motion, vibrations, and electrical circuits. It is also used to optimize and solve problems related to business and finance.
How does trigonometry help in understanding the concept of derivatives?
Trigonometric functions are used to represent and calculate the slope of a curve at any given point, which is the fundamental concept of derivatives in calculus. The derivatives of trigonometric functions, such as cosine and sine, have important applications in physics and engineering.
What are some useful strategies for solving calculus problems involving trigonometry?
Some useful strategies for solving calculus problems involving trigonometry include using trigonometric identities, converting trigonometric functions to their exponential forms, and using the fundamental theorem of calculus. It is also important to have a strong understanding of both calculus and trigonometry concepts and how they relate to each other.