Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(c) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(c)

I.

Question 1.

Find the lengths of subtangent and sub-normal at a point of the curve

Solution:

Equation of the curve is y = b. sin \(\frac{x}{a}\)

\(\frac{dy}{dx}\) = b. cos \(\frac{x}{a}.\frac{1}{a}=\frac{b}{a}\). cos \(\frac{x}{a}\)

Length of the sub-tangent = |\(\frac{y_{1}}{f^{\prime}\left(x_{1}\right)}\)|

= \(\frac{b \cdot \sin \frac{x}{a}}{\frac{b}{a} \cdot \cos \frac{x}{a}}=\left|a \cdot \tan \frac{x}{a}\right|\)

Length of the sub-normal = |y_{1}, f'(x_{1})|

Question 2.

Show that the length of sub normal at any point on the curve xy = a2 varies as the cube of the ordinate of the point.

Solution:

Equation of the curve is xy = a²

y = \(\frac{a^{2}}{x}\)

\(\frac{dy}{dx}=\frac{-a^{2}}{x^{2}}\)

Length of the sub-normal = |y_{1}, f’ (x_{1})|

\(\frac{y_{1}^{3}}{a^{2}}\) ∝ y^{3}_{1} = cube of the ordinate.

Question 3.

Show that at any point (x, y) on the curve y = be^{x/a}, the length of the sub¬tangent is a constant and the length of the sub normal is \(\frac{y^{2}}{a}\).

Solution:

Equation of the curve is y = b.e^{x/a}

\(\frac{a^{2}}{x}\) = b.e^{x/a}.\(\frac{1}{a}=\frac{y}{a}\)

Length of the sub tangent

= |\(\frac{y_{1}}{f^{\prime}\left(x_{1}\right)}\)| = \(\frac{y_{1}}{\left(\frac{y_{1}}{a}\right)}\) = a = constant

Length of the sub-normal

= |y_{1}f'(x_{1})| = |y_{1}\(\frac{y_{1}}{a}\)| = \(\frac{y_{1}^{2}}{a}\).

II.

Question 1.

Find the value of k so that the length of the sub normal at any point on the curve xy^{k} = a^{k + 1} is a constant.

Solution:

Equation of the curve is x.y^{k} = a^{k + 1}

Differentiating w.r.to x

x.k.y^{k-1} \(\frac{dy}{dx}\) + y^{k}.1 = 0

x.k.y^{k-1} \(\frac{dy}{dx}\) = -y^{k}

\(\frac{dy}{dx}=\frac{-y^{k}}{k \cdot x \cdot y^{k-1}}=-\frac{y}{kx}\)

Length of the sub-normal = |y_{1}f'(x_{1}|

Length of the sub-normal is constant at any point on the curve is independent of x_{1} and y_{1}

∴ \(\frac{y_{1}^{k+2}}{k \cdot a^{k+1}}\) is independent of x_{1}, y_{1}

⇒ k + 2 = 0 ⇒ k = -2

Question 2.

At any point t on the curve x = a (t + sin t), y = a (1 – cos t), find the lengths of tangent, normal, sub tangent and sub

normal.

Solution:

Equation of the curve is x = a (t + sin t), y = a (1 – cost)

\(\frac{dx}{dt}\) = a (1 + cos t), \(\frac{dy}{dt}\) = a. sin t

Length of the tangent = |y_{1}|\(\sqrt{1+\frac{1}{\left[f^{\prime}\left(x_{1}\right)\right]^{2}}}\)

Length of the normal = |y_{1}|\(\sqrt{1+\left[f\left(x_{1}\right)\right]^{2}}\)

Length of the sub tangent.

= |a (2 sin t/2. cos t/2|

= |a. sin t|

Length of the sub normal

= |y_{1}.f'(x_{1})| = |a(1 – cos t).\(\frac{\sin t / 2}{\cos t / 2}\)|

= |2a sin² t/2. tan t/2|

= |2a sin² t/2. tan t/2|

Question 3.

Find the lengths of normal and sub normal at a point on the curve y = \(\frac{a}{2}\) (e^{x/a} + e^{-x/a}).

Solution:

Equation of the curve is y = \(\frac{a}{2}\) (e^{x/a} + e^{-x/a})

= a. cosh (\(\frac{x}{a}\))

Length of the normal = |y_{1}|\(\sqrt{1+\left[f\left(x_{1}\right)\right]^{2}}\)

= |a.cosh \(\frac{x}{a}\)|\(\sqrt{1+\sinh ^{2} \frac{x}{a}}\)

= a. cosh \(\frac{x}{a}\). cosh \(\frac{x}{a}\) = a. cosh² \(\frac{x}{a}\)

Length of the sub normal = |y_{1} f'(x_{1})|

Question 4.

Find the lengths of subtangent, sub normal at a point t on the curve x = a (cos t + t sin t), y=a (sin t – t cos t)

Solution:

Equations of the curve are x = a (cos t + t sin t)

\(\frac{dx}{dt}\) = a (- sin t + sin t +1 cost) = at cos t

y = a (sin t – t cos t)

\(\frac{dy}{dt}\) = a (cos t – cos t +1 sin t) = a t sin t

Length of the sub tangent

= |a cot t (sin t – t cos t)|

Length of the sub-normal = |y_{1}.f'(x_{1})|

= |a (sin t – t cos t) tan t|

= |a tan t (sin t – t cos t)|