A particle moves along the curve y=3x^2+x so that that the x-coordinate is increasing at a constant rate of 6 units per second. What is the magnitude (in units per second) of the particle's velocity vector when t at the point (0,0)?
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Answered by
1
Answer:
Given:
y=
3
2
x
3
+1
charge in y-coordinate =2 times (the change in x-coordinate)
dy=2dx
dx
dy
=2
To Find: The point at which
dx
dy
=2 in the eq. y=
3
2
x
3
+1
solution:
consider y=
3
2
x
3
+1
Differentiate w.r.t. 'x'
dx
dy
=2x
2
2x
2
=2 [∴
dx
dy
=2]
x
2
=1
x=±1
If x=1;y=
3
5
x=−1;y=
3
1
∴(1,
3
5
) and (−1,
3
1
) are the required points.
Answered by
0
Answer: Multiple-Choice Questions on Applications of Differential Calculus. 125. 126 ... The equation of the tangent to the curve y=x sin x at the point 6 ... 0 3. D. 2702- €. 21. The speed of the particle is decreasing for . (A) t> 2 (B) <3 (C) allt ... both increase at the constant rate of 2 ft/sec, then the rate, in square feet per second,.
Step-by-step explanation:
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