Trigonometric Calculus Problem Solving Question

In summary, the height of the shadow on the wall is changing at a rate of -7/15 m/s when the person is 4m from the wall. This can be found by using the ratio of the distance the person has traveled to the distance from the spotlight to the wall, and setting it equal to the ratio of the height of the person to the height of the shadow. After implicitly differentiating and substituting the given values, the rate of change can be calculated to be -7/15 m/s.
  • #1
sonic25
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Homework Statement


A spotlight on the ground shines on a wall 14m away. A person of height 2 m walks toward the wall on a direct path between the spotlight and the wall at a rate of 5/3m/s. How fast is the height of the shadow on the wall changing when the person is 4m from the wall?

Homework Equations


Not sure? I think implicit differentiation needs to be employed though.

The Attempt at a Solution


I tried to construct a right-angled triangle to use as a diagram to represent the situation, however this didn't really help as I cannot identify which variables I need to use to answer the problem.

Thanks in advance for any help:)
 
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  • #2
sonic25 said:
I tried to construct a right-angled triangle to use as a diagram to represent the situation, however this didn't really help as I cannot identify which variables I need to use to answer the problem.

Hi sonic,

You are supposed to find how the height of the shadow varies with time right? You know that the height depends on the distance of the person from the source. Just write down the relationship between the two and see how it varies with time.
 
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  • #3
Sunil Simha said:
Hi sonic,

You are supposed to find how the height of the shadow varies with time right? You know that the height depends on the distance of the person from the source. Just write down the relationship between the two and see how it varies with time.
Yeah I'm just not sure as to how the 2m height of the person affects this relationship (if it does at all)?
 
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  • #4
Think of a straight line that goes from the spotlight to the top of the guys head and onto the wall. You will have similar triangles, and then you can construct the ratio.

Hint*: What is the length the man has traveled after time t.
 
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  • #5
xiavatar said:
Think of a straight line that goes from the spotlight to the top of the guys head and onto the wall. You will have similar triangles, and then you can construct the ratio.

Hint*: What is the length the man has traveled after time t.
Ok. Here is my latest attempt at a solution:
I devised a diagram like you said and tried to use a ratio.
Let x be the distance from the wall and h be the height of the shadow above the wall.
tan(theta)=h/14
So, in terms of h: h=14tan(theta)
But tan(theta)=2/(14-x)
Substitute this into h=14tan(theta),
h=14(2/(14-x))
I then implicitly differentiated this with respect to time.
dh/dt=-28((14-x)^(-2))(-dx/dt)
I then substituted the values given in the question:
dh/dt=35/12

Is this correct?
 

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  • #6
Your answer is incorrect. First note that you want the height to be a function of time,## t##. Your diagram is correct, but forget ##x## and ##\theta##. After a time t, how far is the man from the spotlight at time ##t##. Now this will be the length the man is from the spotlight. Now you have two similar triangles. One of these triangles is the man and his distance from the spotlight. And the other triangle composed of the height of the wall and the distance the wall is from the spotlight.
 
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  • #7
xiavatar said:
Now you have two similar triangles. One of these triangles is the man and his distance from the spotlight. And the other triangle composed of the height of the wall and the distance the wall is from the spotlight.
I used the similar triangles and gathered that the ratio of (5/3)t:14 was equal to 2:h.
By rearrangement, h=84/(5t)
Therefore, dh/dt=-84/(5t^2)
Since it takes 6 seconds for the man to reach the point 4m from the wall, I substituted t=6.
So, my end result for dh/dt was -7/15.
Is this the right answer?
 
  • #8
Yes, that is correct.
 
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1. What is trigonometric calculus?

Trigonometric calculus is a branch of mathematics that combines trigonometry and calculus. It involves using trigonometric functions such as sine, cosine, and tangent to solve problems related to calculus.

2. What types of problems can be solved using trigonometric calculus?

Trigonometric calculus can be used to solve a wide range of problems, including finding the slope and area under a curve, calculating the maximum and minimum values of a function, and solving optimization problems involving angles and distances.

3. How is trigonometric calculus different from regular calculus?

Trigonometric calculus is different from regular calculus in that it specifically incorporates trigonometric functions into the equations and solutions. This allows for a more specialized and precise approach to solving certain types of problems.

4. What are some common applications of trigonometric calculus?

Trigonometric calculus has many real-world applications, such as in physics, engineering, and astronomy. It is used to model and analyze periodic phenomena, such as waves and vibrations, and to solve problems involving angles and distances.

5. How can one improve their skills in solving trigonometric calculus problems?

Practice is key in improving skills in solving trigonometric calculus problems. It is important to understand the fundamentals of both trigonometry and calculus, and to work on a variety of problems to develop problem-solving techniques. Additionally, seeking guidance from a tutor or teacher can also be helpful in improving skills.

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