Application of Derivatives Questions

If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.

Solution : Let r be the radius of a sphere and \(\delta\)r be the error in measuring the radius. Then, r = 9 cm and \(\delta\)r = 0.03 cm. Let V be the volume of the sphere. Then, V = \({4\over 3}\pi r^3\)  \(\implies\) \(dV\over dr\) = \(4\pi r^2\) \(\implies\) \(({dV\over dr})_{r = 9}\) = …

If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume. Read More »

Verify Rolle’s theorem for the function f(x) = \(x^2\) – 5x + 6 on the interval [2, 3].

Solution : Since a polynomial function is everywhere differentiable and so continuous also. Therefore, f(x) is continuous on [2, 3] and differentiable on (2, 3). Also, f(2) = \(2^2\) – 5 \(\times\) 2 + 6 = 0 and f(3) = \(3^2\) – 5 \(\times\) 3 + 6 = 0 \(\therefore\) f(2) = f(3) Thus, all …

Verify Rolle’s theorem for the function f(x) = \(x^2\) – 5x + 6 on the interval [2, 3]. Read More »

It is given that for the function f(x) = \(x^3 – 6x^2 + ax + b\) on [1, 3], Rolles’s theorem holds with c = \(2 +{1\over \sqrt{3}}\). Find the values of a and b, if f(1) = f(3) = 0.

Solution : We are given that f(1) = f(3) = 0. \(\therefore\)  \(1^3 – 6 \times 1 + a + b\) = \(3^3 – 6 \times 3^2 + 3a + b\) = 0 \(\implies\)  a + b = 5 and 3a + b = 27 Solving these two equations for a and b, f'(c) is …

It is given that for the function f(x) = \(x^3 – 6x^2 + ax + b\) on [1, 3], Rolles’s theorem holds with c = \(2 +{1\over \sqrt{3}}\). Find the values of a and b, if f(1) = f(3) = 0. Read More »

Find the point on the curve y = cos x – 1, x \(\in\) \([{\pi\over 2}, {3\pi\over 2}]\) at which tangent is parallel to the x-axis.

Solution : Let f(x) = cos x – 1, Clearly f(x) is continous on \([{\pi\over 2}, {3\pi\over 2}]\) and differentiable on \(({\pi\over 2}, {3\pi\over 2})\). Also, f\((\pi\over 2)\) = \(cos {\pi\over 2}\) – 1 = -1 = f\((3\pi\over 2)\). Thus, all the conditions of rolle’s theorem are satisfied. Consequently,there exist at least one point c …

Find the point on the curve y = cos x – 1, x \(\in\) \([{\pi\over 2}, {3\pi\over 2}]\) at which tangent is parallel to the x-axis. Read More »

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

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 »

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

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 »

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

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 »

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

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 »

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

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 »