Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt.
y = √x
(a) Find dy/dt, given x = 16 and dx/dt = 7.
dy/dt =

(b) Find dx/dt, given x = 64 and dy/dt = 8.
dx/dt =
...?

Respuesta :

This problem requires the use of the chain rule: dy / dt = [dy / dx] * [dx / dt]

y = √x => dy / dx = 1 / (2√x)

(a) Find dy/dt, given x = 16 and dx/dt = 7.

dy/dt = [ 1/(2√x) ]  * 7 = [1/(2*4)] * 7 = 7/8

(b) Find dx/dt, given x = 64 and dy/dt = 8.

dx/dt = [dy/dt] /  [dy/dx] = 8 / [1/(2√64) ] = 128

(a) The value of [tex]\frac{dy}{dt}[/tex] is [tex]\boxed{\dfrac{dy}{dt}=\dfrac{7}{8}}[/tex].

(b) The value of [tex]\frac{dx}{dt}[/tex] is [tex]\boxed{\dfrac{dx}{dt}=128}[/tex].

Further explanation:

Given that [tex]x[/tex] and [tex]y[/tex] are both differentiable functions of [tex]t[/tex].

The function is as follows:

[tex]\fbox{\begin\\\ \math y=\sqrt{x}\\\end{minispace}}[/tex]

Derivatives of parametric functions:

The relationship between variable [tex]x[/tex] and variable [tex]y[/tex] in the form [tex]y=f(t)[/tex] and [tex]x=g(t)[/tex] is called as parametric form with [tex]t[/tex] as a parameter.

The derivative of the parametric form is given as  follows:

[tex]\fbox{\begin\\\ \dfrac{dy}{dt}=\dfrac{dy}{dx}\times\dfrac{dx}{dt}\\\end{minispace}}[/tex]  

The above equation is neither explicit nor implicit, therefore a third variable is used.

Part (a):

It is given that [tex]x=16[/tex] and [tex]\frac{dx}{dt}=7[/tex].

To find the value of [tex]\frac{dy}{dt}[/tex] first find the value of [tex]\frac{dy}{dx}[/tex] that is shown below:

[tex]\begin{aligned}y&=\sqrt{x}\\\dfrac{dy}{dx}&=\dfrac{1}{2\sqrt{x}}\end{aligned}[/tex]  

The value of [tex]\frac{dy}{dt}[/tex] is calculated as  follows:

[tex]\begin{aligned}\dfrac{dy}{dt}&=\dfrac{dy}{dx}\times\dfrac{dx}{dt}\\&=\dfrac{1}{2\sqrt{x}}\dfrac{dx}{dt}\end{aligned}[/tex]

Now, substitute the value of [tex]x[/tex] and [tex]\frac{dx}{dt}[/tex] in above equation as shown below:

[tex]\begin{aligned}\dfrac{dy}{dt}&=\dfrac{1}{2\sqrt{16}}\times7\\&=\dfrac{1}{2\times4}\times7\\&=\dfrac{7}{8}\end{aligned}[/tex]  

Therefore, the required value of [tex]\frac{dy}{dt}[/tex] is [tex]\frac{7}{8}[/tex].

Part (b):

It is given that [tex]x=64[/tex] and [tex]\frac{dy}{dt}=8[/tex].

To find the value of [tex]\frac{dx}{dt}[/tex] first find the value of [tex]\frac{dx}{dt}[/tex] that is shown below:

[tex]\begin{aligned}y&=\sqrt{x}\\\dfrac{dy}{dx}&=\dfrac{1}{2\sqrt{x}}\\\dfrac{dx}{dy}&=2\sqrt{x}\end{aligned}[/tex]

The value of [tex]\frac{dx}{dt}[/tex] is calculated as  follows:

[tex]\boxed{\dfrac{dx}{dt}=\dfrac{dx}{dy}\times\dfrac{dy}{dt}}[/tex]

Now, substitute the value of [tex]\frac{dx}{dy}[/tex] and [tex]\frac{dy}{dt}[/tex] in above equation as shown below:

[tex]\begin{aligned}\dfrac{dx}{dt}&=2\sqrt{64}\times 8\\&=2\times8\times8\\&=2\times64\\&=128\end{aligned}[/tex]

Therefore, the required value of [tex]\frac{dx}{dt}[/tex] is [tex]128[/tex].

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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.