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A candle of radius 1cm is floating in a liquid in a cylindrical container of radius 1m.If the candle is burning at the rate of 1.5cm/then the top of the candle will fall at what rate?
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A candle of radius 1cm is floating in a liquid in a cylindrical contai...
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A candle of radius 1cm is floating in a liquid in a cylindrical contai...
The Problem

A candle with a radius of 1 cm is floating in a cylindrical container with a radius of 1 m. The candle is burning at a rate of 1.5 cm per hour. We need to determine at what rate the top of the candle will fall.

Understanding the Problem

To solve this problem, we need to apply related rates, a concept in calculus that deals with the rates at which different variables change in relation to each other. In this case, we are given the rate at which the height of the candle is changing (1.5 cm/hr) and we need to find the rate at which the top of the candle is falling.

Solution

Step 1: Establish Variables and Relationships
Let's denote the height of the candle as h (in cm) and the distance from the top of the candle to the liquid level as d (in cm). We are given the rate at which h is changing, which is dh/dt = -1.5 cm/hr (negative sign indicates the decrease in height).

Step 2: Establish Equations
We know that the volume of the candle is constant because the candle is burning. The volume of a cylinder is given by V = πr^2h, where r is the radius and h is the height. Since the radius of the candle does not change, the volume equation becomes V = πh.

Step 3: Relate Variables
We can relate the variables h and d using the Pythagorean theorem. The radius of the container is 1 m, which is equal to 100 cm. Therefore, according to the Pythagorean theorem, we have d^2 + r^2 = h^2.

Step 4: Differentiate with Respect to Time
To find the rate at which the top of the candle is falling, we need to differentiate the equation d^2 + r^2 = h^2 with respect to time t. Differentiating both sides of the equation, we get 2d(dd/dt) = 2h(dh/dt).

Step 5: Substitute Known Values
We can substitute the known values into the equation. We know that r = 1 cm, dh/dt = -1.5 cm/hr, and we need to find dd/dt.

Step 6: Solve for the Rate
Substituting the known values into the equation, we get 2d(dd/dt) = 2h(dh/dt). Since r = 1 cm, d = 100 cm, and dh/dt = -1.5 cm/hr, we can solve for dd/dt:

2(100)(dd/dt) = 2h(-1.5)
200(dd/dt) = -3h

Since r = 1 cm, we can substitute h = √(d^2 + r^2) into the equation:

200(dd/dt) = -3(√(d^2 + 1^2))

Now we can solve for dd/dt:

dd/dt = (-3/200)√(d^2 + 1^2)

Conclusion

The rate at which the top of the candle falls, dd/dt, is given
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A candle of radius 1cm is floating in a liquid in a cylindrical container of radius 1m.If the candle is burning at the rate of 1.5cm/then the top of the candle will fall at what rate? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A candle of radius 1cm is floating in a liquid in a cylindrical container of radius 1m.If the candle is burning at the rate of 1.5cm/then the top of the candle will fall at what rate? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A candle of radius 1cm is floating in a liquid in a cylindrical container of radius 1m.If the candle is burning at the rate of 1.5cm/then the top of the candle will fall at what rate?.
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