The correct question is;
A particle is moving along the curve
y=√x. As the particle passes through the point (4, 2), its x-coordinate increases at a rate of 4 cm/s. How fast is the distance from the particle to the origin changing at this instant? (Round your answer to three decimal places.)
Answer:
dS/dt ≈ 4.025 cm/s
Step-by-step explanation:
Let the distance from the origin be S and thus, the formula for that is
S = √x² + y²
The given function is y =√x
Let's plug in the S function for y to get;
S = √x² + (√x)²
S = √(x² + x)
Now, let's find the derivative with respect to time;
So;
dS/dt= [(1/2)/√(x² + x)] (2x+1)dx/dt
Let's plug in the value of x = 4 and dX/dt as 4;
dS/dt= [[(1/2)/√(4² + 4)] x [((2(4)+1)]] x 4
dS/dt= [(36/4)/√5]
dS/dt = 9/√5
dS/dt = 4.025 cm/s
Answer:
the answer is 4
Step-by-step explanation:
the answer will be positive
F(x)= √3x + 1
a=8
8 √3 + 1 <—Exact form
14.85649646 <—In decimal form
