(C) Given $W = nw$. Using the product rule for differentiation,we have: $\frac{dW}{dt} = n \frac{dw}{dt} + w \frac{dn}{dt}$ Given $n = 2t^2 + 3$ and $

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(C) Given $W = nw$. Using the product rule for differentiation,we have: $\frac{dW}{dt} = n \frac{dw}{dt} + w \frac{dn}{dt}$ Given $n = 2t^2 + 3$ and $w = t^2 - t + 2$. Calculating the derivatives with respect to $t$: $\frac{dn}{dt} = 4t$ $\frac{dw}{dt} = 2t - 1$ At $t = 1$: $n = 2(1)^2 + 3 = 5$ $w = (1)^2 - 1 + 2 = 2$ $\frac{dn}{dt} = 4(1) = 4$ $\frac{dw}{dt} = 2(1) - 1 = 1$ Substituting these values into the derivative formula: $\frac{dW}{dt} = (5)(1) + (2)(4)$ $\frac{dW}{dt} = 5 + 8 = 13$

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

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The weight $W$ of a certain stock of fish is given by $W = nw,$ where $n$ is the size of the stock and $w$ is the average weight of a fish. If $n$ and $w$ change with time $t$ as $n = 2t^2 + 3$ and $w = t^2 - t + 2,$ then the rate of change of $W$ with respect to $t$ at $t = 1$ is

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$1$
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$8$
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$13$
D
$5$

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Applications of Derivatives

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Rate of Change of Quantities

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The weight $W$ of a certain stock of fish is given by $W = nw,$ where $n$ is the size of the stock and $w$ is the average weight of a fish. If $n$ and $w$ change with time $t$ as $n = 2t^2 + 3$ and $w = t^2 - t + 2,$ then the rate of change of $W$ with respect to $t$ at $t = 1$ is

Similar Questions

If the velocity of a moving particle is directly proportional to the square root of the distance it has covered,what is its acceleration?

The diameter and altitude of a right circular cone,at a certain instant,were found to be $10 \ cm$ and $20 \ cm$ respectively. If its diameter is increasing at a rate of $2 \ cm/s$,then at what rate must its altitude change,in order to keep its volume constant (in $cm/s$)?

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