(a) The value of \(\frac{dy}{dt}\) is \(\boxed{\dfrac{dy}{dt}=\dfrac{7}{8}}\).
(b) The value of \(\frac{dx}{dt}\) is \(\boxed{\dfrac{dx}{dt}=128}\).
Further explanation:
Given that \(x\) and \(y\) are both differentiable functions of \(t\).
The function is as follows:
\(\fbox{\begin\\\ \math y=\sqrt{x}\\\end{minispace}}\)
Derivatives of parametric functions:
The relationship between variable \(x\) and variable \(y\) in the form \(y=f(t)\) and \(x=g(t)\) is called as parametric form with \(t\) as a parameter.
The derivative of the parametric form is given as follows:
\(\fbox{\begin\\\ \dfrac{dy}{dt}=\dfrac{dy}{dx}\times\dfrac{dx}{dt}\\\end{minispace}}\)
The above equation is neither explicit nor implicit, therefore a third variable is used.
Part (a):
It is given that \(x=16\) and \(\frac{dx}{dt}=7\).
To find the value of \(\frac{dy}{dt}\) first find the value of \(\frac{dy}{dx}\) that is shown below:
\(\begin{aligned}y&=\sqrt{x}\\\dfrac{dy}{dx}&=\dfrac{1}{2\sqrt{x}}\end{aligned}\)
The value of \(\frac{dy}{dt}\) is calculated as follows:
\(\begin{aligned}\dfrac{dy}{dt}&=\dfrac{dy}{dx}\times\dfrac{dx}{dt}\\&=\dfrac{1}{2\sqrt{x}}\dfrac{dx}{dt}\end{aligned}\)
Now, substitute the value of \(x\) and \(\frac{dx}{dt}\) in above equation as shown below:
\(\begin{aligned}\dfrac{dy}{dt}&=\dfrac{1}{2\sqrt{16}}\times7\\&=\dfrac{1}{2\times4}\times7\\&=\dfrac{7}{8}\end{aligned}\)
Therefore, the required value of \(\frac{dy}{dt}\) is \(\frac{7}{8}\).
Part (b):
It is given that \(x=64\) and \(\frac{dy}{dt}=8\).
To find the value of \(\frac{dx}{dt}\) first find the value of \(\frac{dx}{dt}\) that is shown below:
\(\begin{aligned}y&=\sqrt{x}\\\dfrac{dy}{dx}&=\dfrac{1}{2\sqrt{x}}\\\dfrac{dx}{dy}&=2\sqrt{x}\end{aligned}\)
The value of \(\frac{dx}{dt}\) is calculated as follows:
\(\boxed{\dfrac{dx}{dt}=\dfrac{dx}{dy}\times\dfrac{dy}{dt}}\)
Now, substitute the value of \(\frac{dx}{dy}\) and \(\frac{dy}{dt}\) in above equation as shown below:
\(\begin{aligned}\dfrac{dx}{dt}&=2\sqrt{64}\times 8\\&=2\times8\times8\\&=2\times64\\&=128\end{aligned}\)
Therefore, the required value of \(\frac{dx}{dt}\) is \(128\).
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Answer details:
Grade: Senior school
Subject: Mathematics
Chapter: Derivatives
Keywords: dy/dt, dx/dt, y=rootx, derivatives, parametric form, implicit, explicit, function, differentiable, x, y, t, x=16, x=64, dy/dx, dx/dy, differentiation.
