Finding equation of a tangent and normal line to the given curve at the given point?

Question

could you solve these and please explain how? 

1) y=ln(e^x+e^2x) at (0,ln2) 

2) sqrt(x)+sqrt(y)=3 at (4,1) 

3) x^(2/3)+y^(2/3)=1 at (1/2^(3/2), 1/2^(3/2))

Answer 1

• 1).

The curve is y = ln(e^x + e^2x) and the point is (0, ln 2).

Differentiate the curve with respect to x.

y ' = [1/(e^x + e^2x)][e^x + 2e^2x]

y ' = [e^x + 2e^2x]/(e^x + e^2x)

y ' = e^x[1 + 2e^x]/e^x(1 + e^x)

⇒ y ' = [1 + 2e^x]/[1 + e^x].

When, x = 0, y ' = [1 + 2e⁰]/[1 + e⁰] = (1 + 2)/(1 + 1) = 3/2.

y ' = 3/2.

This is the slope (m ) of the tangent line to the implicit curve at (0, ln 2).

 

Slope - intercept form line equation is y = mx + b, where m is slope and b is y - intercept.

Now the tangent line equation is y = (3/2)x  + b.

Find the y - intercept by substituting the the point in the tangent line equation say (x, y) = (0, ln 2).

ln 2 = (3/2)0  + b

b = ln 2 -  0

⇒ b = ln 2.

The tangent line equation  is y = (3/2)x + ln 2.

 

The normal line and tangent are perpendecular to each other.

Since the slopes of perpendecular lines are negative reciprocals the slope of nolmal line through the point  (0, ln 2) is - 2/3.

Slope (m) = - 2/3.

Now, the normal line equation is y = (- 2/3)x + b.

Find the y - intercept by substituting the the point in the normal line equation say (x, y) = (0, ln 2).

ln 2 = (- 2/3)(0) + b

b = ln 2 + 0

⇒ b = ln 2.

The normal line equation  is y = (- 2/3)x + ln 2.

 

• 2).

The curve is √x + √y = 3 and the point is (4, 1).

Differentiate the curve with respect to x.

(1/2√x) + (1/2√y)y ' = 0

⇒ y ' = - √(y/x).

When, x = 4 and y = 1, y ' = - √(1/4) = - 1/2.

y ' = - 1/2.

This is the slope (m ) of the tangent line to the implicit curve at (4, 1).

 

Slope - intercept form line equation is y = mx + b, where m is slope and b is y - intercept.

Now the tangent line equation is y = (- 1/2)x  + b.

Find the y - intercept by substituting the the point in the tangent line equation say (x, y) = (4, 1).

1 = (- 1/2)(4) + b

b = 1 + 2

⇒ b = 3.

The tangent line equation  is y = (- 1/2)x + 3.

 

The normal line and tangent are perpendecular to each other.

Since the slopes of perpendecular lines are negative reciprocals the slope of nolmal line through the point  (4, 1) is 2.

Slope (m) = 2.

Now, the normal line equation is y = 2x + b.

Find the y - intercept by substituting the the point in the normal line equation say (x, y) = (4, 1).

1 = (2)(4) + b

b = 1 - 8

⇒ b = - 7.

The normal line equation  is y = 2x - 7.

Answer 2

Contd........

• 3).

The curve is and the point is .

Differentiate the curve with respect to x.

.

When, x = 0.3535 and y = 0.3535, .

y ' = - 1.

This is the slope (m ) of the tangent line to the implicit curve at (0.3535, 0.3535).

 

Slope - intercept form line equation is y = mx + b, where m is slope and b is y - intercept.

Now the tangent line equation is y = - x  + b.

Find the y - intercept by substituting the the point in the tangent line equation say (x, y) = (0.3535, 0.3535).

0.3535 = (- 0.3535) + b

b = 0.3535 + 0.3535

⇒ b = 0.707.

The tangent line equation  is y = - x + 0.707.

 

The normal line and tangent are perpendecular to each other.

Since the slopes of perpendecular lines are negative reciprocals the slope of nolmal line through the point  (0.3535, 0.3535) is 1.

Slope (m) = 1.

Now, the normal line equation is y = x + b.

Find the y - intercept by substituting the the point in the normal line equation say (x, y) = (0.3535, 0.3535).

  0.3535 = (0.3535) + b

b = 0.3535 - 0.3535

⇒ b = 0.

The normal line equation  is y = x.