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ANEK [815]
3 years ago
8

Assume that

Mathematics
1 answer:
kap26 [50]3 years ago
7 0

The value of \frac{dy}{dt} is 396

Explanation:

It is given that x=x(t) and y=y(t)

Also, given that y=x^3+1 and \frac{dx}{dt} =33 when x=22

Now, we need to determine the value of \frac{dy}{dt} when x=2

To determine the value of \frac{dy}{dt} when x=2, we need to differentiate the function y=x^3+1 with respect to \frac{dy}{dt}

Thus, we have,

\frac{dy}{dt}=3x^2(\frac{dx}{dt} )+0

Simplifying, we get,

\frac{dy}{dt}=3x^2(\frac{dx}{dt} )

Substituting x=2 and \frac{dx}{dt} =33 , we get,

\frac{dy}{dt}=3(2)^2(33 )

Squaring the term, we have,

\frac{dy}{dt}=3(4)(33 )

Multiplying the terms, we get,

\frac{dy}{dt}=396

Thus, the value of \frac{dy}{dt} is 396

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