Question

Assume that

x equals x left parenthesis t right parenthesisx=x(t)

and

y equals y left parenthesis t right parenthesisy=y(t).

Let

y equals x cubed plus 1y=x3+1

and

StartFraction dx Over dt EndFractiondxdt​=33

when

xequals=22.

Find

StartFraction dy Over dt EndFractiondydt

when

xequals=2

StartFraction dy Over dt EndFractiondydtequals=

Answer 1

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 $396$