TS EAMCET · Maths · Application of Derivatives
A particle moves along the curve \(y=x^2+2 x\). Then, the point on the curve such that \(x\) and \(y\) coordinates of the particle change with the same rate is
- A \((1,3)\)
- B \(\left(\frac{1}{2}, \frac{5}{2}\right)\)
- C \(\left(-\frac{1}{2},-\frac{3}{4}\right)\)
- D \((-1,-1)\)
Answer & Solution
Correct Answer
(C) \(\left(-\frac{1}{2},-\frac{3}{4}\right)\)
Step-by-step Solution
Detailed explanation
Since, \(x\) and \(y\) particles both have the same rate of change \(\therefore \quad \frac{d x}{d t}=\frac{d y}{d t}\) ...(i) Given equation of curve is \( y=x^2+2 x \) On differentiating both sides w.r.t. \(t\), we get \( \frac{d y}{d t}=(2 x+2) \frac{d x}{d t} \)…
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