Question

Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt.

Answer 1

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

Answer 2

(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 $128$.

Learn more:

1. A problem on trigonometric ratio brainly.com/question/9880052

2. A problem on function brainly.com/question/3412497

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.