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 = |y1, f'(x1)|

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 = |y1, f’ (x1)|

$$\frac{y_{1}^{3}}{a^{2}}$$ ∝ y31 = cube of the ordinate.

Question 3.
Show that at any point (x, y) on the curve y = bex/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.ex/a
$$\frac{a^{2}}{x}$$ = b.ex/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
= |y1f'(x1)| = |y1$$\frac{y_{1}}{a}$$| = $$\frac{y_{1}^{2}}{a}$$.

II.

Question 1.
Find the value of k so that the length of the sub normal at any point on the curve xyk = ak + 1 is a constant.
Solution:
Equation of the curve is x.yk = ak + 1
Differentiating w.r.to x
x.k.yk-1 $$\frac{dy}{dx}$$ + yk.1 = 0
x.k.yk-1 $$\frac{dy}{dx}$$ = -yk
$$\frac{dy}{dx}=\frac{-y^{k}}{k \cdot x \cdot y^{k-1}}=-\frac{y}{kx}$$
Length of the sub-normal = |y1f'(x1|

Length of the sub-normal is constant at any point on the curve is independent of x1 and y1
∴ $$\frac{y_{1}^{k+2}}{k \cdot a^{k+1}}$$ is independent of x1, y1
⇒ 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 = |y1|$$\sqrt{1+\frac{1}{\left[f^{\prime}\left(x_{1}\right)\right]^{2}}}$$

Length of the normal = |y1|$$\sqrt{1+\left[f\left(x_{1}\right)\right]^{2}}$$

Length of the sub tangent.

= |a (2 sin t/2. cos t/2|
= |a. sin t|
Length of the sub normal
= |y1.f'(x1)| = |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 = $$\frac{a}{2}$$ (ex/a + e-x/a).
Solution:
Equation of the curve is y = $$\frac{a}{2}$$ (ex/a + e-x/a)
= a. cosh ($$\frac{x}{a}$$)
Length of the normal = |y1|$$\sqrt{1+\left[f\left(x_{1}\right)\right]^{2}}$$
= |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 = |y1 f'(x1)|

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 = |y1.f'(x1)|
= |a (sin t – t cos t) tan t|
= |a tan t (sin t – t cos t)|