Maths Questions

What is the parametric equation of ellipse ?

Solution : The equation x = acos\(\theta\) & y = bsin\(\theta\) together represent the parametric equation of ellipse \({x_1}^2\over a^2\) + \({y_1}^2\over b^2\) = 1, where \(\theta\) is a parameter. Note that if P(\(\theta\)) = (acos\(\theta\), bsin\(\theta\)) is on the ellipse then ; Q(\(\theta\)) = (acos\(\theta\), bsin\(\theta\)) is on auxilliary circle. A circle described on …

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Differentiate \(x^{sinx}\) with respect to x.

Solution : Let y = \(x^{sinx}\). Then, Taking log both sides, log y = sin x.log x \(\implies\) y = \(e^{sin x.log x}\) By using logarithmic differentiation, On differentiating both sides with respect to x, we get \(dy\over dx\) = \(e^{sin x.log x}\)\(d\over dx\)(sin x.log x) \(\implies\) \(dy\over dx\) = \(x^{sin x}{log x {d\over dx}(sin …

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If y = \(\sqrt{sinx + \sqrt{sinx + \sqrt{sinx + ……. to \infty}}}\), find \(dy\over dx\).

Solution : Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as y = \(\sqrt{sin x + y}\) Squaring on both sides, \(\implies\)  \(y^2\)  = sin x + y By using differentiation of infinite series, Differentiating both sides with respect to x, 2y \(dy\over …

If y = \(\sqrt{sinx + \sqrt{sinx + \sqrt{sinx + ……. to \infty}}}\), find \(dy\over dx\). Read More »

Find \(dy\over dx\) where x = a{cos t + \({1\over 2} log tan^2 {t\over 2}\)} and y = a sin t

Solution : We have, x = a{cos t + \({1\over 2} log tan^2 {t\over 2}\)} and y = a sin t \(\implies\) x = a{cos t + \({1\over 2} \times 2 log tan{t\over 2}\)} and y = a sin t \(\implies\) x = a{cos t + {\(log tan{t\over 2}\)} and y = a sin t …

Find \(dy\over dx\) where x = a{cos t + \({1\over 2} log tan^2 {t\over 2}\)} and y = a sin t Read More »

Find the determinant of A = \(\begin{bmatrix} 3 & -2 & 4 \\ 1 & 2 & 1 \\ 0 & 1 & -1 \end{bmatrix}\).

Solution : | A | = \(\begin{vmatrix} 3 & -2 & 4 \\ 1 & 2 & 1 \\ 0 & 1 & -1 \end{vmatrix}\) By using 3×3 determinant formula, \(\implies\) | A | = \(3\begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix}\) – \((-2)\begin{vmatrix} 1 & 1 \\ 0 & -1 \end{vmatrix}\) + …

Find the determinant of A = \(\begin{bmatrix} 3 & -2 & 4 \\ 1 & 2 & 1 \\ 0 & 1 & -1 \end{bmatrix}\). Read More »