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Ori was given this problem: The radius r(t) of the base of a cone is increasing at a rate of 10 meters per second. The height h(t) of the cone is fixed at 6 meters. At a certain instant t_(0), the radius is 1 meter. What is the rate of change of the volume V(t) of the cone at that instant? Which equation should Ori use to solve the problem? Choose 1 answer: (A) V(t)=pi[r(t)]^(2)+2pi*r(t)*h(t) (B) V(t)=pi[r(t)]^(2)*h(t) (C) V(t)=pi[r(t)]^(2)+pi*r(t)sqrt([r(t)]^(2)+[h(t)]^(2)) (D) V(t)=(pi[r(t)]^(2)*h(t))/(3)

Ori was given this problem:\newlineThe radius r(t) r(t) of the base of a cone is increasing at a rate of 1010 meters per second. The height h(t) h(t) of the cone is fixed at 66 meters. At a certain instant t0 t_{0} , the radius is 11 meter. What is the rate of change of the volume V(t) V(t) of the cone at that instant?\newlineWhich equation should Ori use to solve the problem?\newlineChoose 11 answer:\newline(A) V(t)=π[r(t)]2+2π⋅r(t)⋅h(t) V(t)=\pi[r(t)]^{2}+2 \pi \cdot r(t) \cdot h(t) \newline(B) V(t)=π[r(t)]2⋅h(t) V(t)=\pi[r(t)]^{2} \cdot h(t) \newline(C) V(t)=π[r(t)]2+π⋅r(t)[r(t)]2+[h(t)]2 V(t)=\pi[r(t)]^{2}+\pi \cdot r(t) \sqrt{[r(t)]^{2}+[h(t)]^{2}} \newline(D) V(t)=π[r(t)]2⋅h(t)3 V(t)=\frac{\pi[r(t)]^{2} \cdot h(t)}{3}

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Q. Ori was given this problem:\newlineThe radius r(t) r(t) of the base of a cone is increasing at a rate of 1010 meters per second. The height h(t) h(t) of the cone is fixed at 66 meters. At a certain instant t0 t_{0} , the radius is 11 meter. What is the rate of change of the volume V(t) V(t) of the cone at that instant?\newlineWhich equation should Ori use to solve the problem?\newlineChoose 11 answer:\newline(A) V(t)=π[r(t)]2+2π⋅r(t)⋅h(t) V(t)=\pi[r(t)]^{2}+2 \pi \cdot r(t) \cdot h(t) \newline(B) V(t)=π[r(t)]2⋅h(t) V(t)=\pi[r(t)]^{2} \cdot h(t) \newline(C) V(t)=π[r(t)]2+π⋅r(t)[r(t)]2+[h(t)]2 V(t)=\pi[r(t)]^{2}+\pi \cdot r(t) \sqrt{[r(t)]^{2}+[h(t)]^{2}} \newline(D) V(t)=π[r(t)]2⋅h(t)3 V(t)=\frac{\pi[r(t)]^{2} \cdot h(t)}{3}
  1. Volume Function Derivation: The volume VV of a cone is given by the formula V=13πr2hV = \frac{1}{3}\pi r^2 h, where rr is the radius and hh is the height. Since the height h(t)h(t) is fixed at 66 meters, we can write the volume as a function of time tt using the radius r(t)r(t).
  2. Radius Function Expression: Given that r(t)r(t) is increasing at a rate of 1010 meters per second, we can express r(t)r(t) as r(t)=r0+10tr(t) = r_0 + 10t, where r0r_0 is the initial radius. At the instant t0t_0, the radius r(t0)r(t_0) is 11 meter, so r0=1r_0 = 1.
  3. Rate of Change Calculation: To find the rate of change of the volume V(t)V(t) at the instant t0t_0, we need to differentiate the volume function with respect to time tt. This requires using the correct formula for the volume of a cone.
  4. Substitution of Values: The correct formula for the volume of a cone is V(t)=13π[r(t)]2hV(t) = \frac{1}{3}\pi[r(t)]^2h. This matches with option (D) V(t)=π[r(t)]2⋅h(t)3V(t) = \frac{\pi[r(t)]^2 \cdot h(t)}{3}, since h(t)h(t) is constant and equals 66 meters.
  5. Derivative Calculation: Now we will differentiate the volume function V(t)V(t) with respect to time tt using the chain rule. The derivative of V(t)V(t) with respect to tt is dVdt=(13)π⋅2r(t)⋅drdt⋅h\frac{dV}{dt} = (\frac{1}{3})\pi \cdot 2r(t) \cdot \frac{dr}{dt} \cdot h, where drdt\frac{dr}{dt} is the rate of change of the radius with respect to time.
  6. Final Rate of Change: Substitute the given values into the derivative to find the rate of change of the volume at the instant t0t_0. We have h=6h = 6 meters, drdt=10\frac{dr}{dt} = 10 meters/second, and r(t0)=1r(t_0) = 1 meter.\newlinedVdt=(13)π×2×1\frac{dV}{dt} = (\frac{1}{3})\pi \times 2 \times 1 meter ×10\times 10 meters/second ×6\times 6 meters
  7. Final Rate of Change: Substitute the given values into the derivative to find the rate of change of the volume at the instant t0t_0. We have h=6h = 6 meters, drdt=10\frac{dr}{dt} = 10 meters/second, and r(t0)=1r(t_0) = 1 meter.dVdt=(13)π×2×1\frac{dV}{dt} = (\frac{1}{3})\pi \times 2 \times 1 meter ×10\times 10 meters/second ×6\times 6 metersPerform the calculation to find the rate of change of the volume at the instant t0t_0.dVdt=(13)π×2×1×10×6\frac{dV}{dt} = (\frac{1}{3})\pi \times 2 \times 1 \times 10 \times 6dVdt=(13)π×120\frac{dV}{dt} = (\frac{1}{3})\pi \times 120h=6h = 600 cubic meters per second

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