Question

find the of change of the distance D from the origin of a point moving on the graph off …dx/dt= 9 units per second. Write the exact answer do not round

Answer 1

Given:

D is start origin that mean (0,0) moving with graph :

`f(x)=x^2`

It move x then y is

`\begin{gathered} f(x)=x^2 \\ y=x^2 \end{gathered}`

Point is (2,4)

Distance is:

`D=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2_{}}`

Where,

`\begin{gathered} x_2=x \\ x_1=0 \\ y_2=x^2 \\ y_1=0 \end{gathered}`

So distance is:

`\begin{gathered} D=\sqrt[]{(x-0)^2+(x^2-0)^2} \\ D=\sqrt[]{x^2+x^4} \end{gathered}`
`\begin{gathered} (dD)/(\differentialDt t)=\frac{d(\sqrt[]{x^2+x^4})}{\differentialDt t} \\ =\frac{1}{2\sqrt[]{x^2+x^4}}(2x+4x^3)(dx)/(\differentialDt t) \end{gathered}`
`\begin{gathered} \text{when x=2} \\ (dx)/(\differentialDt t)=9 \end{gathered}`
`\begin{gathered} =\frac{1}{2\sqrt[]{(2^2+2^4)}}((2*2)+(4*2^3))*9 \\ =(324)/(6.92820) \\ =46.7653 \end{gathered}`