MHB Find the rate of change of the distance between the origin and a moving point

[karush]

Find the rate of change of the distance between the origin and a moving point on the graph of
$$y=\sin{x} \text{ if } \frac{dx}{dt}=\frac{2 \text{cm}}{\text {sec}}$$

 

[MarkFL]

You are asked to find the rate of change of the distance between a point on the curve and the origin, so you need to use the distance formula. Let's let $D(t)$ be this distance as a function of time $t$, and so we may state:

(1) $$D^2(t)=x^2(t)+y^2(t)$$

where we may parametrize the sinusoid as follows:

$$x(t)=2t$$

$$y(t)=\sin(2t)$$

Differentiate (1) with respect to $t$. Since you are not actually given an $x$-value at which to evaluate the rate of change $D'(t)$, it will be a function of $t$.

 

[karush]

$$ \displaystyle D^2(t)=x^2(t)+y^2(t) \Rightarrow D^2(t)=4t^2+\sin^2{2t}$$ $$ \frac{dD}{dt}2D=8t + 4 \sin{2t}\cos{2t} $$ $$ \frac{dD}{dt}D=4t + 2 \sin{2t}\cos{2t} $$ $$ \frac{dD}{dt}D=4(2)+2\sin{(2\cdot2)}\cos{(2\cdot2)}\approx 9\frac{cm}{sec} $$ don't think I understand the function of t in this??

 

[MarkFL]

You have not solved for $$\frac{dD}{dt}$$. I parametrized $x$ and $y$ since we are given $$\frac{dx}{dt}$$. Does the problem state to find $D'(2)$?

 

[karush]

so you mean that

$D=\sqrt{4t^2+\sin^2{(2t)}}$

then take $\displaystyle\frac{d}{dt}$

 

[MarkFL]

You could do that, but it is simpler to differentiate implicitly:

$$D^2(t)=x^2(t)+y^2(t)$$

$$2D(t)\frac{dD}{dt}=2x(t)\frac{dx}{dt}+2y(t)\frac{dy}{dt}$$

Hence:

$$\frac{dD}{dt}=\frac{x(t)\frac{dx}{dt}+y(t)\frac{dy}{dt}}{D(t)}$$

 

[karush]

$$D^2(t)=x^2(t)+y^2(t)$$

$$2D(t)\frac{dD}{dt}=2x(t)\frac{dx}{dt}+2y(t)\frac{dy}{dt}$$

Hence:

$$\frac{dD}{dt}=\frac{x(t)\frac{dx}{dt}+y(t)\frac{dy}{dt}}{D(t)}$$

$$\frac{dD}{dt}=
\frac{x(t)\frac{dx}{dt}+y(t)\frac{dy}{dt}}
{D(t)}$$

$$x(t)=2t $$
$$\displaystyle y(t)=\sin(2t)$$
$$t=2$$
$$\frac{dx}{dt}=\frac{2 \text{cm}}{\text {sec}}$$

$$\displaystyle\frac{dD}{dt}=\frac{(2\cdot 2 )\frac{2 \text{cm}}{\text {sec}}+\sin{(2\cdot2)}\frac{dy}{dt}}
{D(t)}$$

what is $D(t)$ here?

 

[MarkFL]

$$\frac{dD}{dt}=
\frac{x(t)\frac{dx}{dt}+y(t)\frac{dy}{dt}}
{D(t)}$$

$$x(t)=2t $$
$$\displaystyle y(t)=\sin(2t)$$
$$t=2$$
$$\frac{dx}{dt}=\frac{2 \text{cm}}{\text {sec}}$$

$$\displaystyle\frac{dD}{dt}=\frac{(2\cdot 2 )\frac{2 \text{cm}}{\text {sec}}+\sin{(2\cdot2)}\frac{dy}{dt}}
{D(t)}$$

what is $D(t)$ here?

From the parametrization, we have:

$$D(t)=\sqrt{(2t)^2+\sin^2(2t)}$$

 

[karush]

$$
\displaystyle\frac{dD}{dt}
=\frac{(2\cdot 2 )\frac{2 \text{cm}}{\text {sec}}+2\cos{(2\cdot2)}}
{\displaystyle \sqrt{(2\cdot2)^2+\sin^2(2\cdot2)}}
\approx 1.64 cm/sec
$$

its stab in dark ...

 

[MarkFL]

$$
\displaystyle\frac{dD}{dt}
=\frac{(2\cdot 2 )\frac{2 \text{cm}}{\text {sec}}+2\cos{(2\cdot2)}}
{\displaystyle \sqrt{(2\cdot2)^2+\sin^2(2\cdot2)}}
\approx 1.64 cm/sec
$$

its stab in dark ...

That looks correct to me for $t=2$.