(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

Gas is being pumped into a spherical balloon at the rate of $30 \ ft^3 / \text{min}$. The rate at which the radius increases when it reaches the value $15 \ ft$,is:

$A$ spherical iron ball of radius $10 \, cm$ is coated with a layer of ice of uniform thickness that melts at a rate of $50 \, cm^3/min$. When the thickness of the ice is $5 \, cm$,find the rate at which the thickness of the ice decreases.

The distance $s$ moved by a particle in time $t$ is given by $s = f(t) = t^3 - 6t^2 + 9t$,where $s$ is in metres and $t$ is in seconds. The velocity of the particle at $t = 2 \ s$ is: (in $m/s$)

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