Application of Derivatives Questions

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 = ${4\over 3}\pi r^3$  $\implies$ $dV\over dr$ = $4\pi r^2$ $\implies$ $({dV\over dr})_{r = 9}$ = […]

Solution : Let y = f(x), x = 3 and x + $\delta x$. Then, $\delta x$.= 0.02. For x = 3, we get y = f(3) = 45 Now, y = f(x) $\implies$ y = $3x^2 + 5x + 3$ $\implies$ $dy\over dx$ = 6x + 5 $\implies$  $({dy\over dx})_{x = 3}$ = 23

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

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

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

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

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 $\theta$ = $\pi\over 2$

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

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

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