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Water leaking onto a floor forms a circular pool. The area of the pool increases at a rate of 16picm2/min16\,\text{picm}^{2}/\text{min}. How fast is the radius of the pool increasing when the radius is 7cm7\,\text{cm}?

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Q. Water leaking onto a floor forms a circular pool. The area of the pool increases at a rate of 16picm2/min16\,\text{picm}^{2}/\text{min}. How fast is the radius of the pool increasing when the radius is 7cm7\,\text{cm}?
  1. Area Formula: The area AA of a circle is given by the formula A=πr2A = \pi r^2, where rr is the radius of the circle. We are given that the area of the pool increases at a rate of 16picm2/min16 \, \text{picm}^2/\text{min}, which is dAdt=16picm2/min\frac{dA}{dt} = 16 \, \text{picm}^2/\text{min}. We need to find the rate at which the radius rr is increasing, which is drdt\frac{dr}{dt}, when the radius is 7cm7\,\text{cm}.
  2. Chain Rule Application: To find drdt\frac{dr}{dt}, we will use the chain rule from calculus, which relates the rates of change of two related quantities. Differentiating both sides of the area formula with respect to time tt, we get dAdt=d(πr2)dt\frac{dA}{dt} = \frac{d(\pi r^2)}{dt}. Applying the chain rule, we get dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}.
  3. Substitution and Calculation: We can now solve for drdt\frac{dr}{dt} by substituting the given rate of change of the area and the radius at the moment we are interested in. We have dAdt=16pi cm2/min\frac{dA}{dt} = 16 \, \text{pi cm}^2/\text{min} and r=7cmr = 7\,\text{cm}. Plugging these values into the equation, we get 16=2π7drdt16 = 2\pi \cdot 7 \cdot \frac{dr}{dt}.
  4. Isolating drdt\frac{dr}{dt}: Solving for drdt\frac{dr}{dt}, we divide both sides of the equation by 2π×72\pi \times 7 to isolate drdt\frac{dr}{dt}. This gives us drdt=16(2π×7)\frac{dr}{dt} = \frac{16}{(2\pi \times 7)}.
  5. Final Calculation: Now we perform the calculation: drdt=162π7=1614π=87π\frac{dr}{dt} = \frac{16}{2 \cdot \pi \cdot 7} = \frac{16}{14\pi} = \frac{8}{7\pi} cm/min.

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