Respuesta :
Answer:7 cm/s
Explanation:
Given
Particle move along curve
[tex]y=\sqrt{1+x^3}[/tex]
As it reaches the (2,3) its y coordinate is increasing at 14 cm/s
Differentiating y w.r.t time
[tex]\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
Now at (2,3)
[tex]\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s[/tex]
The speed at which the x-coordinate of the point is changing at the given instant is; 7 cm/s
What is the instant velocity?
We are given the parametric curve in which the particle moves as;
y = √(1 + x³)
We are told that as it reaches the point (2, 3), the y-coordinate is increasing at a rate of 14 cm/s.
Thus, dy/dt = 14 cm/s
Thus, we differentiate with respect to time;
dy/dt = [3x²/(2√(1 + x³))] * (dx/dt)
Plugging in the coordinate value of x and value of dy/dt, we have;
14 = [3(2)²/(2√(1 + 2³))] * (dx/dt)
14 = (12/6) * (dx/dt)
dx/dt = 14/2
dx/dt = 7 cm/s
Read more about Instant Velocity at; https://brainly.com/question/2234298
