If V = 4/3 πr³, at what rate in cubic units is V increasing when r = 10 and dr/dt = 0.01?
Cubic units measure the volume of a three-dimensional object, representing the space it occupies. A cubic unit is the volume of a cube with sides of one unit length, such as 1 cm³ or 1 m³. It is widely used in geometry, physics, and engineering to quantify capacities, densities and material quantities in real-world applications.
Class 12 Maths Chapter 6 focuses on Applications of Derivatives for CBSE Exam 2024-25. It covers rate of change of quantities increasing and decreasing functions tangents and normals maxima and minima. These topics help solve real-life problems in physics economics and engineering while enhancing problem-solving skills and understanding of mathematical applications in various fields.
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To find the rate at which the volume V is increasing, we need to differentiate the volume formula V = (4/3) π r³ with respect to time t.
dV/dt = 4 π r² (dr/dt)
Now, plug in the values: r = 10 and (dr/dt) = 0.01:
dV/dt = 4 π (10)² (0.01)
dV/dt = 4 π × 100 × 0.01 = 4 π
Thus, the rate at which the volume is increasing is 4π cubic units.
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