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

## Find the approximate value of f(3.02), where f(x) = $$3x^2 + 5x + 3$$.

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 …

## 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 …

## 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 …

## 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 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 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 $$\theta$$ = $$\pi\over 2$$ …

## 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 …

## 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 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 …