Solution : We have, x = \(a sin^3 t\), y = \(b cos^3 t\) \(\implies\)  \(dx\over dt\) = \(3a sin^2t cos t\)  and, \(dy\over dt\)  = \(-3b cos^2t sin t\) \(\therefore\)   \(dy\over dx\) = \(dy/dt\over dx/dt\) = \(-b\over a\) \(cos t\over sin t\) So, the equation of the tangent at the point ‘t’ is y […]

Solution : The equation of the given curve is y = \(2x^2 + 3 sin x\)                   ……….(i) Putting x = 0 in (i), we get y = 0. So, the point of contact is (0, 0). Now, y = \(2x^2 + 3 sin x\) Differentiating with respect

Solution : The equation of the two curves are xy = 6          …….(i) and, \(x^2 y\) = 12            …………(ii) from (i) , we obtain y = \(6\over x\). Putting this value of y in (ii), we obtain \(x^2\) \((6\over x)\) = 12 \(\implies\) 6x = 12

Solution : Solving the curves simultaneously we get points of intersection as (1, 1) and (0, 0). At (1, 1) for first curve \(2y({dy\over dx})_1\) = 1  \(\implies\)  \(m_1\) = \(1\over 2\) & for second curve 2x = \(({dy\over dx})_2\) \(\implies\)  \(m_2\) = 2 \(m_1m_2\) = -1 at (1, 1). But at (0, 0) clearly

Solution : \(x^2\) = 32y  \(\implies\)  \(dy\over dx\) = \(x\over 16\)  \(\implies\)  \(y^2\) = 4x \(\implies\)  \(dy\over dx\) = \(2\over y\) \(\therefore\)  at  (16, 8), \((dy\over dx)_1\) = 1, \((dy\over dx)_2\) = \(1\over 4\) So, required angle = \(tan^{-1}({1 – {1\over 4}\over 1 + 1({1\over 4})})\) = \(tan^{-1}({3\over 5})\) Similar Questions Find the equations of

Solution : We have,  f(x) = \(-x^2 – 2x + 15\) \(\implies\) f'(x) = -2x – 2 = -2(x + 1) for f(x) to be increasing, we must have f'(x) > 0 -2(x + 1) > 0 \(\implies\) x + 1 < 0 \(\implies\) x < -1 \(\implies\) x \(\in\) \((-\infty, -1)\). Thus f(x) is

Solution : We have, \(f(\theta)\) = \({4sin \theta\over 2 + cos\theta} – \theta\) \(\implies\) \(f'(\theta)\) = \((2 + cos\theta)(4 cos\theta) + 4 sin^2\theta\over (2 + cos\theta)^2\) – 1 \(\implies\) \(f'(\theta)\)  = \(8 cos\theta + 4\over (2 + cos\theta)^2\) – 1 \(\implies\) \(f'(\theta)\) = \(4\cos\theta – cos^2\theta\over (2 + cos\theta)^2\) \(\implies\) \(f'(\theta)\) = \(cos\theta(4 – cos\theta)\over

Solution : We have, f(x) = \(x^3 – 3x^2 + 3x – 100\) \(\implies\)  f'(x) = \(3x^2 – 6x + 3\) = \(3(x – 1)^2\) Now, x \(\in\) R \(\implies\)  \((x – 1)^2\)  \(\ge\)  0  \(\implies\)  f'(x)  \(\ge\) 0. Thus, f'(x) \(\ge\) 0 for all x \(\in\) R. Hence, f(x) is increasing on R. Similar

Solution : We have, f(x) = sin 3x \(\therefore\)   f'(x) = 3cos 3x Now,  0 < x < \(pi\over 2\)   \(\implies\)  0 < 3x < \(3\pi\over 2\) Since cosine function is positive in first quadrant and negative in the second and third quadrants. Therefore, we consider the following cases. Case 1 : When 0 <

Solution : f(x) = \(x^4\over 12\) – \(5x^3\over 6\) + \(3x^2\) + 7. f'(x) = \(x^3\over 3\) – \(5x^2\over 2\) + 6x f”(x) = \(x^2\) – 5x + 6 Since, f”(x) = 0 at point of inflection. \(\implies\) \(x^2\) – 5x + 6 = 0 \(\implies\) x = 2 and x = 3 Hence, points