TS Inter 1st Year Maths 1B Solutions Chapter 10 Applications of Derivatives Ex 10(d)

Students must practice this TS Inter 1st Year Maths 1B Study Material Chapter 10 Applications of Derivatives Ex 10(d) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(d)

I. Find the angle between the curves given below. (March 2014) (E.Q.)

Question 1.
x + y + 2 = 0, x2 + y2 – 10y = 0
Answer:
x + y + 2 = 0 ⇒ x = – (y + 2)
x2 + y2 – 10y = 0
⇒ (y + 2)2 + y2 – 10y = 0
⇒ y2 + 4y + 4 + y2 – 10y = 0
⇒ 2y2 – 6y + 4 = 0
⇒ y2 – 3y + 2 = 0
⇒ (y – 2) (y – 1) = 0
⇒ y = 1 (or) y = 2
x = – (y + 2)
y = 1 ⇒ x = -(1 + 2) = -3
y = 2 ⇒ x = -(2 + 2) = -4
The points of intersection at P (- 3, 1) and Q (- 4, 2)
Equation of curve is x + y – 10y = 0
Differentiating w.r.to ‘x’
TS Inter 1st Year Maths 1B Solutions Chapter 10 Applications of Derivatives Ex 10(d) 1

TS Inter 1st Year Maths 1B Solutions Chapter 10 Applications of Derivatives Ex 10(d)

Question 2.
y2 = 4x, x2 + y2 = 5. (E.Q.) (May 2007)
Answer:
The given equations of curves are y2 = 4x and x2 + y2 = 5
Eliminating y, we get x2 + 4x = 5
⇒ x2 + 4x – 5 = 0
⇒ (x – 1) (x + 5) = 0
⇒ x = 1 or – 5
From y2 = 4x and x = 1
⇒ y2 = 4 ⇒ y = ±2
But when x = -5, y is imaginary and not real
Points of Intersection are
P(1, 2) and Q(1, -2)
From the equation y2 = 4x
⇒ 2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 4
⇒ \(\frac{d y}{d x}=\frac{z}{y}\) …….(1)
At P(1, 2), slope of first curve from (1)
m1 = \(\frac{2}{2}\) = 1
and m2 = slope of second curve from (2)
= \(\frac{-1}{2}\)
TS Inter 1st Year Maths 1B Solutions Chapter 10 Applications of Derivatives Ex 10(d) 2

Question 3.
x2 + 3y = 3, x2 – y2 + 25 = 0 (E.Q.)
Answer:
x2 + 3y = 3 ……….(1),
x2 – y2 + 25 = 0 ………….(2)
are the equations of given curves.
From (1) x2 = 3 – 3y
From (2), 3 – 3y – y2 + 25 = 0
⇒ – y2 – 3y + 28 = 0
⇒ y2 + 3y – 28 = 0
⇒ (y + 7)(y – 4) = 0
⇒ y = 4 or y = -7
If y = 4 then x2 = 3 – 12 = -9
Values of x are not real.
If y = – 7 then x2 = 3 + 21
⇒ x2 = 24 = ± 2√6
∴ Points of intersection are
P (2√6 , -7) and Q (-2√6 , -7)
From (1), 2x + 3 \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=\frac{-2 x}{3}\)
Slope of the tangent at P (2√6 , -7) to the curve (1) is m1 = \(\frac{-2(2 \sqrt{6})}{3}=\frac{-4 \sqrt{6}}{3}\)
TS Inter 1st Year Maths 1B Solutions Chapter 10 Applications of Derivatives Ex 10(d) 3

TS Inter 1st Year Maths 1B Solutions Chapter 10 Applications of Derivatives Ex 10(d)

Question 6.
y2 = 8x, 4x2 + y2 = 32
Answer:
y = 8x ….(1)
4x2 + y2 = 32
⇒ 4x2 + 8x – 32 = 0
⇒ x2 + 2x – 8 = 0
⇒ (x + 4) (x – 2) = 0
⇒ x = 2 or x = – 4
When x = 2, y2 = 16 ⇒ y = ± 4
∴ Points of intersection are P (2, 4) and Q (2, – 4)
From (1), 2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 8 ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{\mathrm{y}}\)
TS Inter 1st Year Maths 1B Solutions Chapter 10 Applications of Derivatives Ex 10(d) 4

Question 7.
x2y = 4, y(x2 + 4) = 8.
Answer:
x2y = 4 ………………(1)
y(x2 + 4) = 8 …………(2)
From (1), x2 = \(\frac{4}{y}\)

Substitute x2 value in equation (2)
∴ y(\(\frac{4}{y}\)+4) = 8 ⇒ y\(\left(\frac{4+4 y}{y}\right)\) = 8
⇒ 4 + 4y = 8 ⇒ y = 1
∴ x2 = \(\frac{4}{y}\)
⇒ x2 = 4
⇒ x = ±2
∴ Points of intersection are P(2, 1), Q(-2, 1)
From (1)
TS Inter 1st Year Maths 1B Solutions Chapter 10 Applications of Derivatives Ex 10(d) 5

TS Inter 1st Year Maths 1B Solutions Chapter 10 Applications of Derivatives Ex 10(d)

Question 8.
Show that the curves 6x2 – 5x + 2y = 0 and 4x2 + 8y2 = 3 touch each other at \(\left(\frac{1}{2}, \frac{1}{2}\right)\). (E.Q)
Answer:
The given curves are 6x2 – 5x + 2y = 0 ………..(1)
and 4x2 + 8y2 = 3 ………..(2)
TS Inter 1st Year Maths 1B Solutions Chapter 10 Applications of Derivatives Ex 10(d) 6
Slopes of two curves at \(\left(\frac{1}{2}, \frac{1}{2}\right)\) are equal and hence the given curves touch each other at \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

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