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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives1 Mark
MCQ
If $\text{x}=\text{f}(\text{t})\cos\text{t}-\text{f}(\text{t})\sin\text{t}\ \text{and}\ \text{y}=\text{f}(\text{t})\sin\text{t}+\text{f}(\text{t})\cos\text{t},$ then $\Big(\frac{\text{dx}}{\text{dt}}\Big)^2+\Big(\frac{\text{dy}}{\text{dt}}\Big)^2=$
  • A
    $\text{f}(\text{t})-\text{f}(\text{t})$
  • B
    $\{\text{f}(\text{t})-\text{f}(\text{t})\}^2$
  • ✓
    $\{\text{f}(\text{t})+\text{f}(\text{t})\}^2$
  • D
    $\text{None of these}$
✓

Answer

Correct option: C.
$\{\text{f}(\text{t})+\text{f}(\text{t})\}^2$
Here,
$\text{x}=\text{f}(\text{t})\cos\text{t}-\text{f}'(\text{t})\sin\text{t}$
$\text{and}\ \text{y}=\text{f}(\text{t})\sin\text{t}+\text{f}'(\text{t})\cos\text{t}$
$ \Rightarrow\frac{\text{dx}}{\text{dt}}=\text{f}'(\text{t})\cos\text{t}-\text{f}(\text{t})\sin\text{t}-\text{f}''(\text{t})\sin\text{t}-\text{f}'(\text{t})\cos\text{}\text{t}$
$ \text{and}\ \frac{\text{dy}}{\text{dt}}=\text{f}'(\text{t})\sin\text)\cos\text{t}+\text{f}'{t}+\text{f}(\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\text{f}(\text{t})\sin\text{t}-\text{f}''(\text{t})\sin\text{t}$
$ \text{and}\ \frac{\text{dy}}{\text{dt}}=\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}$
Thus
$\Big(\frac{\text{dx}}{\text{dt}}\Big)^2+\Big(\frac{\text{dy}}{\text{dt}}\Big)^2=\{-\text{f}(\text{t})\sin\text{t}\}^2+\{\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}\Big\}^2$
$=\{\text{f}(\text{t})\sin\text{t}+\text{f}''(\text{t})\sin\text{t}\}^2+\{\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}\}^2$
$=\sin^2\text{t}\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2+\cos^2\text{t}\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2$
$=\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2(\sin^2\text{t}+\cos^2\text{t})$
$=\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2$

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