Mathemerize

Solution : We have, x = \(a cos^3\theta\), y = \(a sin^3\theta\) \(\implies\)  \(dx\over d\theta\) = \(-3 a cos^2\theta sin\theta\),  \(dy\over d\theta\) = \(3 a sin^2\theta cos\theta\) Now, \(dy\over dx\)  = \(dy/d\theta\over dx/d\theta\) \(\implies\)  \(dy\over dx\) = -\(tan\theta\) \(\therefore\)  Slope of the normal at any point on the curve = \(-1\over dy/dx\) = \(cot\theta\) Hence, the …

Find the slope of the normal to the curve x = \(a cos^3\theta\), y = \(a sin^3\theta\) at \(\theta\) = \(\pi\over 4\). Read More »

Solution : We have, x = 1 – \(a sin\theta\), y = \(b cos^2\theta\) \(\implies\)  \(dx\over d\theta\) = \(-a cos\theta\)  and \(dy\over d\theta\) = \(-2b cos\theta sin\theta\) \(\therefore\)  \(dy\over dx\) = \(dy/d\theta\over dx/d\theta\) = \(2b\over a\) \(sin\theta\) \(\implies\)  \(dy\over dx\)  at \(\pi\over 2\) = \(2b\over a\) Hence, Slope of normal at \(\theta\) = \(\pi\over 2\) …

Find the slope of normal to the curve x = 1 – \(a sin\theta\), y = \(b cos^2\theta\) at \(\theta\) = \(\pi\over 2\). Read More »

Solution : The equation of the curve is y = \(2x^3 – 3\) Differentiating with respect to x, we get \(dy\over dx\) = \(6x^2\) Now, \(m_1\) = (Slope of the tangent at x = 2) = \(({dy\over dx})_{x = 2}\) = \(6 \times (2)^2\) = 24 and, \(m_2\) = (Slope of the tangent at x …

Show that the tangents to the curve y = \(2x^3 – 3\) at the points where x =2 and x = -2 are parallel. Read More »

Solution : The equation of the given curve is y = \(-5x^2 + 6x + 7\) \(\implies\) \(dy\over dx\) = -10x + 6 \(\implies\) \(({dy\over dx})_{(1/2, 35/4)}\) = \(-10\over 4\) + 6 = 1 The required equation at (1/2, 35/4) is y – \(35\over 4\) = \(({dy\over dx})_{(1/2, 35/4)}\) \((x – {1\over 2})\) \(\implies\) y …

Find the equation of the tangent to curve y = \(-5x^2 + 6x + 7\)  at the point (1/2, 35/4). Read More »

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 …

Find the equations of the tangent and the normal at the point ‘t’ on the curve x = \(a sin^3 t\), y = \(b cos^3 t\). Read More »

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 …

Find the equation of the normal to the curve y = \(2x^2 + 3 sin x\) at x = 0. Read More »

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 …

Find the angle between the curves xy = 6 and \(x^2 y\) =12. Read More »

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 …

Check the orthogonality of the curves \(y^2\) = x and \(x^2\) = y. Read More »

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 …

The angle of intersection between the curve \(x^2\) = 32y and \(y^2\) = 4x at point (16, 8) is Read More »

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 …

Find the interval in which f(x) = \(-x^2 – 2x + 15\) is increasing or decreasing. Read More »