Answer:
The rate increases at 398mi/h
Step-by-step explanation:
Given
[tex]\frac{dx}{dt} = 460mi/h[/tex] --- rate when [tex]y = 2[/tex] --- distance
The given parameters is represented with the attached figure
Using Pythagoras theorem [take the attached image as a point of reference]
[tex]y^2 = x^2 + 1^2[/tex]
[tex]y^2 = x^2 + 1[/tex]
Differentiate with respect to time (t):
[tex]2y \frac{dy}{dt} = 2x \frac{dx}{dt}[/tex]
Divide both sides by 2
[tex]y \frac{dy}{dt} = x \frac{dx}{dt}[/tex] --- (1)
When y = 2:
[tex]y^2 = x^2 + 1^2[/tex] --- Pythagoras Theorem
[tex]2^2 = x^2 + 1[/tex]
[tex]4 = x^2 + 1[/tex]
Collect like terms
[tex]x^2 = 4 -1[/tex]
[tex]x^2 = 3[/tex]
Solve for x
[tex]x = \sqrt 3[/tex]
So:
[tex]y \frac{dy}{dt} = x \frac{dx}{dt}[/tex]
Substitute: [tex]x = \sqrt 3[/tex], [tex]\frac{dx}{dt} = 460mi/h[/tex] and [tex]y = 2[/tex]
[tex]2 * \frac{dy}{dt} = \sqrt 3 * 460[/tex]
[tex]2 * \frac{dy}{dt} = 460\sqrt 3[/tex]
Divide by 2
[tex]\frac{dy}{dt} = 230\sqrt 3[/tex]
[tex]\frac{dy}{dt} = 230*1.7321[/tex]
[tex]\frac{dy}{dt} = 398.383[/tex]
[tex]\frac{dy}{dt} \approx 398[/tex]
