Question

Find the rate of change of the distance from the origin of a moving point on the graph

Answer 1

EXPLANATION

Given the function f(x) = x^2 and the point x=2

Now, we need a formula for the Distance as follows:

$\frac{dD}{dt}=(\frac{dD}{dx})\cdot(\frac{dx}{d\text{ t}})$

We need to compute dD/dx, this means the derivative of the distance between the points (0,0) and (2,x^2) .

The distance must be computed using the distance formula:

$D=\sqrt[]{(x-0)^2+(x^2-0)^2}$

Adding terms and computing the powers:

$D=\sqrt[]{x^2+x^4}$

Rearranging terms:

$D=(x^4+x^2)^{\frac{1}{2}}$

Then, we must apply the first derivative of this expression by applying the chain rule as follows:

$\frac{df(u)}{dx}=\frac{df}{du}\cdot\frac{du}{dx}$$f=u^{\frac{1}{2}},\text{ u=}x^4+x^2$$=\frac{d}{du}\mleft(u^{\frac{1}{2}}\mright)\frac{d}{dx}\mleft(x^4+x^2\mright)$$=\frac{1}{2u^{\frac{1}{2}}}\frac{d}{dx}\mleft(x^4+x^2\mright)$$\mathrm{Substitute\: back}\: u=x^4+x^2$$=\frac{1}{2\left(x^4+x^2\right)^{\frac{1}{2}}}\frac{d}{dx}\mleft(x^4+x^2\mright)$

Applying the derivatives:

$=\frac{1}{2\left(x^4+x^2\right)^{\frac{1}{2}}}\mleft(4x^3+2x\mright)$

Simplifying:

$=\frac{2x^2+1}{\left(x^2+1\right)^{\frac{1}{2}}}$

Plugging in x=2 into the expression:

$\frac{dD}{d\text{x}}_{(2)}=\frac{2*2^2+1}{(2^2+1)^{\frac{1}{2}}}=\frac{2*4+1}{(4+1)^{\frac{1}{2}}}=\frac{8+1}{(5)^{\frac{1}{2}}}=\frac{9}{\sqrt[]{5}}$

Finally, as we already know that dx/dt = 7, we must replace terms in the expression:

$\frac{dD}{dt}=(\frac{dD}{dx})\cdot(\frac{dx}{d\text{ t}})$$\frac{dD}{d\text{ t}}=\frac{9}{\sqrt[]{5}}\cdot7$

Multiplying numbers:

$\frac{dD}{d\text{ t}}=\frac{63}{\sqrt[]{5}}=\frac{63\sqrt[]{5}}{5}=28.17$

The solution is 28.17