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

find the of change of the distance D from the origin of a point moving on the graph off dxdt 9 units per second Write the exact answer do not round class=

Respuesta :

Given:

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

[tex]f(x)=x^2[/tex]

It move x then y is

[tex]\begin{gathered} f(x)=x^2 \\ y=x^2 \end{gathered}[/tex]

Point is (2,4)

Distance is:

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

Where,

[tex]\begin{gathered} x_2=x \\ x_1=0 \\ y_2=x^2 \\ y_1=0 \end{gathered}[/tex]

So distance is:

[tex]\begin{gathered} D=\sqrt[]{(x-0)^2+(x^2-0)^2} \\ D=\sqrt[]{x^2+x^4} \end{gathered}[/tex][tex]\begin{gathered} \frac{dD}{\differentialDt t}=\frac{d(\sqrt[]{x^2+x^4})}{\differentialDt t} \\ =\frac{1}{2\sqrt[]{x^2+x^4}}(2x+4x^3)\frac{dx}{\differentialDt t} \end{gathered}[/tex][tex]\begin{gathered} \text{when x=2} \\ \frac{dx}{\differentialDt t}=9 \end{gathered}[/tex][tex]\begin{gathered} =\frac{1}{2\sqrt[]{(2^2+2^4)}}((2\times2)+(4\times2^3))\times9 \\ =\frac{324}{6.92820} \\ =46.7653 \end{gathered}[/tex]