Question:

Find the equation of all lines having slope -1 that are tangents to the curve y=1​/x−1,x=1.

Answer:

Step 1: Find the first derivative of the given equation.

y’ = -1/(x-1)^2

Step 2: Find the slope (m) of the tangent line.

m = -1

Step 3: Find the point of tangency (x, y).

x = 1 y = 1/0 = undefined

Step 4: Use the point-slope form of the equation of a line.

y - y1 = m(x - x1)

y - undefined = -1(x - 1)

y = -x + 1

Question:

Find the points on the curve x^2+y^2−2x−3=0 at which the tangents are parallel to the x-axis.

Answer:

Step 1: Rewrite the equation given in standard form: x^2 + y^2 - 2x - 3 = 0

Step 2: Differentiate the equation with respect to x: 2x - 2 = 0

Step 3: Solve for x: x = 1

Step 4: Substitute x = 1 in the original equation: y^2 - 2 - 3 = 0

Step 5: Solve for y: y = ±√5

Step 6: The points on the curve at which the tangents are parallel to the x-axis are (1, √5) and (1, -√5).

Question:

Find the equations of the tangent and normal to the given curve at the indicated point: y=x4−6x^3+13x^2−10x+5 at (1,3)

Answer:

Answer:

Equation of Tangent:

Slope of the tangent at (1,3) = 4(1)3 − 6(1)2(3) + 13(1)(3)2 − 10(3) + 5 = 4 - 18 + 39 - 30 + 5 = 10

Therefore, equation of the tangent at (1,3) is y - 3 = 10(x - 1) => y = 10x - 7

Equation of Normal:

Slope of the normal at (1,3) = -1/10

Therefore, equation of the normal at (1,3) is y - 3 = -1/10(x - 1) => y = -1/10x + 13/10

Question:

Find the equation of the tangent line to the curve y=x^2−2x+7 which is. (a) parallel to the line 2x−y+9=0. (b) perpendicular to the line 5y−15x=13.

Answer:

(a) Parallel to the line 2x−y+9=0

The equation of the tangent line must have the same slope as the given line.

Slope of the given line, m = 2

Therefore, the equation of the tangent line is y = 2x - 9

(b) Perpendicular to the line 5y−15x=13

The equation of the tangent line must have a slope that is the negative reciprocal of the given line.

Slope of the given line, m = -3/5

Therefore, the equation of the tangent line is y = -3/5x + 13/5

Question:

Find the equations of the tangent and normal to the parabola y^2=4ax at the point (at^2,2at).

Answer:

1. Find the equation of the parabola: y^2=4ax

2. Find the coordinates of the given point: (at^2,2at)

3. Find the derivative of the parabola: y’= (2/4a)x

4. Substitute the coordinates of the given point into the derivative: y’= (2/4a)(at^2)

5. Simplify: y’= (1/2a)t^2

6. Find the equation of the tangent line: y-2at=(1/2a)t^2(x-at^2)

7. Find the equation of the normal line: y-2at=-(1/2a)t^2(x-at^2)

Question:

Find the equations of all lines having slope 0 which are tangent to the curve y=1​/x^2−2x+3.

Answer:

Step 1: Find the derivative of y=1/x2−2x+3.

y’ = -2x/x^3 + 2/x^2

Step 2: Set the derivative equal to zero.

-2x/x^3 + 2/x^2 = 0

Step 3: Factor the equation.

-2x(x^2 - 1) + 2(x^2 - 1) = 0

Step 4: Solve for x.

x^2 - 1 = 0

x = ±1

Step 5: Substitute the x-values into the original equation.

y = 1/x^2 - 2x + 3

y = 1/1^2 - 2(1) + 3

y = 1 - 2 + 3

y = 2

Therefore, the equations of the lines having slope 0 which are tangent to the curve y=1/x^2−2x+3 are x = 1, y = 2 and x = -1, y = 2.

Question:

The line y=x+1 is a tangent to the curve y^2=4x at the point. A (1,2) B (2,1) C (1,4) D (2,2)

Answer:

Step 1: To determine which point is a tangent to the curve, we need to determine the slope of the line y=x+1. The slope of the line is m=1.

Step 2: To determine the slope of the curve y^2=4x at the point, we need to take the derivative of the equation. The derivative of the equation is y’=2y/4x.

Step 3: To determine the slope of the curve at the point, we need to substitute the coordinates of the point into the equation. For point A (1,2), the slope of the curve at the point is m=1. For point B (2,1), the slope of the curve at the point is m=0.5. For point C (1,4), the slope of the curve at the point is m=2. For point D (2,2), the slope of the curve at the point is m=1.

Step 4: Since the slope of the line y=x+1 is m=1, the only point that is a tangent to the curve is point D (2,2).

Question:

Find the equations of the tangent and normal to the given curve at the indicated point: y=x^3 at (1,1)

Answer:

Equation of the tangent:

The equation of the tangent line at (1,1) can be found by taking the derivative of the equation y=x^3.

y’ = 3x^2

The equation of the tangent line is given by y - 1 = 3(x - 1).

Equation of the normal:

The equation of the normal line at (1,1) can be found by taking the perpendicular slope of the equation y=x^3.

The perpendicular slope is -1/3.

The equation of the normal line is given by y - 1 = -1/3(x - 1).

Question:

Find the slope of the normal to the curve x=1−asinθ,y=bcos2θ at θ=π​/2.

Answer:

1. Differentiate the equation x=1−asinθ with respect to θ: dx/dθ = -a*cosθ

2. Differentiate the equation y=bcos2θ with respect to θ: dy/dθ = -2b*sin2θ

3. Find the slope of the normal line: m = -(dy/dθ)/(dx/dθ) m = -(-2bsin2θ)/(-acosθ) m = 2ab*sin2θ/cosθ

4. Substitute θ=π/2: m = 2ab*sin(π/2)/cos(π/2) m = 2ab

Question:

Find the slope of the tangent to curve y=x^3−x+1 at the point whose x-coordinate is 2.

Answer:

1. Find the equation of the tangent line:

y = 3x^2 - 1

2. Find the slope of the tangent line:

m = 6

Question:

Find points on the curve x^2​/9+y^2​/16=1 at which the tangents are Parallel to x-axis are (a,±b).Find a+b

Answer:

Given equation is x^2​/9+y^2​/16=1

Step 1: Differentiating both sides with respect to x,

2x/9 + 0 = 0

Step 2: Solving for x,

x = 0

Step 3: Substituting x = 0 in the given equation,

0 + y^2/16 = 1

Step 4: Solving for y,

y^2 = 16

Step 5: Taking square root of both sides,

y = ±4

Step 6: Substituting the value of y in the given equation,

x^2/9 + (±4)^2/16 = 1

Step 7: Solving for x,

x^2 = 9

Step 8: Taking square root of both sides,

x = ±3

Thus, the points at which the tangents are parallel to x-axis are (a,±b) where a = 3 and b = 4.

Therefore, a + b = 3 + 4 = 7

Question:

Find points at which the tangent to the curve y=x^3−3x^2−9x+7 is parallel to the x-axis.

Answer:

1. Find the derivative of y = x^3−3x^2−9x+7

dy/dx = 3x^2 - 6x - 9

2. Set the derivative equal to 0 in order to find the critical points

3x^2 - 6x - 9 = 0

3. Solve the equation to find the critical points

3x^2 - 6x - 9 = 0

(3x + 3)(x - 3) = 0

x = -3 and x = 3

4. Plug the critical points into the derivative to determine if they are maxima or minima

dy/dx = 3(-3)^2 - 6(-3) - 9 = -36

dy/dx = 3(3)^2 - 6(3) - 9 = 0

Since the derivative of the critical point at x = 3 is equal to 0, the tangent is parallel to the x-axis at that point.

Question:

Find a point on the curve y=(x−2)^2 at which the tangent is parallel to the chord joining the points (2,0) and (4,4).

Answer:

1. Find the equation of the line joining the points (2,0) and (4,4).

The equation of a line joining two points (x1, y1) and (x2, y2) is given by: y = mx + c where m = (y2-y1)/(x2-x1)

Therefore, the equation of the line joining the points (2,0) and (4,4) is: y = 2x

2. Find the equation of the tangent at any point (x,y) on the curve y=(x−2)^2.

The equation of the tangent at any point (x,y) on the curve y=(x−2)^2 is given by: dy/dx = 2(x-2)

3. Set the equation of the line and the equation of the tangent equal to each other and solve for x.

2x = 2(x-2) 2x = 2x - 4 4 = 4

Therefore, the point on the curve y=(x−2)^2 at which the tangent is parallel to the chord joining the points (2,0) and (4,4) is (4,4).

Question:

Find points on the curve 9x^2​+16y^2​=1 at which the tangents are (i) Parallel to x-axis . (ii) Parallel to y-axis.

Answer:

(i) Parallel to x-axis

The equation of the line parallel to the x-axis is y = 0.

Substituting this into the equation of the curve, we get:

9x^2 + 16(0)^2 = 1

9x^2 = 1

x^2 = 1/9

x = ±1/3

Therefore, the points on the curve at which the tangents are parallel to the x-axis are (1/3, 0) and (-1/3, 0).

(ii) Parallel to y-axis

The equation of the line parallel to the y-axis is x = 0.

Substituting this into the equation of the curve, we get:

9(0)^2 + 16y^2 = 1

16y^2 = 1

y^2 = 1/16

y = ±1/4

Therefore, the points on the curve at which the tangents are parallel to the y-axis are (0, 1/4) and (0, -1/4).

Question:

Find the points on the curve y=x^3 at which the slope of the tangent is equal to the y coordinate of the point.

Answer:

1. Set the equation for the slope of the tangent line equal to the y-coordinate of the point: m = y

2. Take the derivative of the equation y=x^3 to get the equation for the slope of the tangent line: m = 3x^2

3. Set the equation for the slope of the tangent line equal to the equation for the y-coordinate of the point: 3x^2 = x^3

4. Solve for x: x = ±√(1/3)

Question:

For the curve y=4x^3−2x^5, find all the points at which the tangents passes through the origin.

Answer:

1. Set y = 0: 0 = 4x^3 − 2x^5
2. Factor the equation: 0 = 2x^5(2−x^2)
3. Set each factor equal to zero: 2 = 0 2 − x^2 = 0
4. Solve for x: 2 = 0 → x = 0 2 − x^2 = 0 → x = ±√2
5. Therefore, the points at which the tangents passes through the origin are (0,0) and (±√2, 0).

Question:

Find the equation of the normal to the curve y=x^3+2x+6 which are parallel to the line x+14y+4=0.

Answer:

Answer:

Step 1: Find the derivative of the equation y=x^3+2x+6.

dy/dx = 3x^2 + 2

Step 2: Find the slope of the normal line.

Slope of normal line = -1/Slope of original line

Slope of original line = -(1/14)

Therefore, slope of normal line = 14

Step 3: Find the equation of the normal line using the point-slope form of the line equation.

y - y1 = m(x - x1)

y - (x^3+2x+6) = 14(x - x1)

Step 4: Substitute the equation of the normal line into the given line equation x+14y+4=0.

x + 14(y - (x^3+2x+6)) + 4 = 0

x + 14y - 14x^3 - 28x - 84 + 4 = 0

Step 5: Simplify the equation.

-14x^3 - 28x + 88 = 0

Question:

Find the equation of all lines having slope 2 which are tangents to the curve y=1​/x−3,x=3.

Answer:

Step 1: Take the derivative of the curve y=1/x-3 with respect to x.

dy/dx = -1/(x-3)^2

Step 2: Set the derivative equal to the slope of the line, which is 2.

-1/(x-3)^2 = 2

Step 3: Solve for x.

-1/(2(x-3)^2) = 1

x-3 = ±√(1/2)

x = 3 ±√(1/2)

Step 4: Substitute the value of x into the equation of the curve.

y = 1/(3 ±√(1/2)) - 3

Step 5: Solve for y.

y = 1/(3 ±√(1/2)) - 3

y = -2 ±√(1/2)

Step 6: The equation of the lines having slope 2 which are tangents to the curve y=1/x-3,x=3 is given by:

y = -2 ±√(1/2)

Question:

Find the slope of the normal to the curve x=acos^3θ,y=asin^3θ at θ=π​/4.

Answer:

Step 1: Find the derivative of the function x=acos^3θ,y=asin^3θ

dx/dθ = -3acos^2θsinθ dy/dθ = 3asin^2θcosθ

Step 2: Find the slope of the normal to the curve

Slope of the normal = -1/m, where m = dy/dx

Step 3: Substitute the values of dy/dx and θ=π/4 in the equation

Slope of the normal = -1/(-3acos^2(π/4)sin(π/4))

Step 4: Simplify the equation

Slope of the normal = -1/(-3(1/2)sin(π/4))

Step 5: Solve for the slope of the normal

Slope of the normal = -2/√2

Question:

Find the point on the curve y=x^3−11x+5 at which the tangent is y=x−11.

Answer:

1. Set y=x^3−11x+5 equal to y=x−11
2. x^3−12x+16=0
3. Factor the equation to get (x-4)(x-4)(x-4)=0
4. The point on the curve is x=4

Question:

Find the equation of the tangent line to the curve y=x^2−2x+7 which is perpendicular to the line 5y−15x=13.

Answer:

1. Find the slope of the line 5y−15x=13. Slope = -15/5

2. Find the slope of the tangent line. The slope of the tangent line is the derivative of the given equation, which is 2x-2.

3. Find the equation of the line perpendicular to the given line. The equation of the line perpendicular to the given line is y = (15/5)x - 13/5.

4. Find the equation of the tangent line. The equation of the tangent line is y = -(2x - 2)(15/5)x + (2x - 2)(-13/5) + 7. Simplifying, the equation of the tangent line is y = -30x + 39/5 + 7.

Question:

Find the equations of the tangent and normal to the given curve at the indicated point: y=x^2 at (0,0).

Answer:

1. First, we need to find the slope of the curve at (0,0). To do this, we can take the derivative of y=x^2, which is y’=2x. Since x=0 at (0,0), the slope of the curve at (0,0) is y’=2(0)=0.

2. Now, we can use the slope to find the equation of the tangent line. The equation of a line in slope-intercept form is y=mx+b, where m is the slope and b is the y-intercept. Since the slope of the tangent line is 0, the equation of the tangent line is y=0x+b, or y=b. Therefore, the equation of the tangent line at (0,0) is y=0.

3. Finally, we can use the slope to find the equation of the normal line. The equation of a line in slope-intercept form is y=mx+b, where m is the slope and b is the y-intercept. Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, the slope of the normal line is -1/0, or undefined. Therefore, the equation of the normal line at (0,0) is undefined.

Question:

Find the equations of the tangent and normal to the given curves at the indicated points: (i) y=x^4−6x^3+13x^2−10x+5 at (0,5). (ii) y=x^4−6x^3+13x^2−10x+5 at (1,3) (iii) y=x^3 at (1,1) (iv) y=x^2 at (0,0) (v) x=cost,y=sint at t=π/4

Answer:

(i) Tangent: y - 5 = 4(x - 0)(x^2 - 6x + 13) Normal: y - 5 = -1/4(x - 0)(3x^2 - 12x + 26)

(ii) Tangent: y - 3 = 4(x - 1)(x^2 - 5x + 8) Normal: y - 3 = -1/4(x - 1)(3x^2 - 10x + 16)

(iii) Tangent: y - 1 = 3(x - 1)(x^2 - 2x + 1) Normal: y - 1 = -1/3(x - 1)(2x^2 - 3x + 2)

(iv) Tangent: y - 0 = 2(x - 0)(x - 0) Normal: y - 0 = -1/2(x - 0)(2x - 0)

(v) Tangent: y - sin(π/4) = cos(π/4)(x - cos(π/4)) Normal: y - sin(π/4) = -1/cos(π/4)(x - cos(π/4))

Question:

Find the slope of the tangent to the curve y=x−1​/x−2,x=2 at x=10.

Answer:

Step 1: Find the equation of the tangent line at x=10.

The equation of the tangent line is y = mx + b, where m is the slope and b is the y-intercept.

Step 2: Calculate the slope of the tangent line at x=10.

The slope of the tangent line is m = (f(x+h) - f(x))/h, where f(x) is the equation of the curve and h is a small value.

Substituting the values:

m = (f(10+h) - f(10))/h

m = ((x-1)/(x-2)|x=10+h - (x-1)/(x-2)|x=10)/h

m = ((10+h-1)/(10+h-2) - (10-1)/(10-2))/h

m = (h-1)/(h-2)

Step 3: Set h = 0.

m = (0-1)/(0-2)

m = -1/ -2

Step 4: Simplify the slope.

m = 1/2

Question:

Find the slope of the tangent to the curve y=3x^4−4x at x = 4.

Answer:

1. Rewrite the equation in the form y=f(x): y=3x^4−4x

2. Find the derivative of the equation: f’(x)=12x^3−4

3. Evaluate the derivative at x=4: f’(4)=12(4)^3−4=192−4=188

4. The slope of the tangent to the curve at x=4 is 188.

Question:

Find the equation of the normal at the point (am^2,am^3 ) for the curve ay^2=x^3

Answer:

Answer: Step 1: Calculate the slope of the curve at the given point (am^2,am^3).

Slope at the point (am^2,am^3) = (3x^2)/(2y) = (3(am^2))/(2(am^3)) = 3/(2am)

Step 2: Calculate the equation of the normal at the point (am^2,am^3).

Equation of the normal at the point (am^2,am^3) = y-(am^3) = -(3/(2am))(x-(am^2)) = -3/(2am)x + 3am^2/2 + am^3

Question:

Show that the tangents to the curve y=7x^3+11 at the points where x=2 and x=−2 are parallel.

Answer:

Step 1: Find the derivative of y=7x^3+11

The derivative of y=7x^3+11 is dy/dx = 21x^2

Step 2: Find the slope of the tangent lines at x = 2 and x = -2

At x = 2, the slope of the tangent line is dy/dx = 21(2)^2 = 84.

At x = -2, the slope of the tangent line is dy/dx = 21(-2)^2 = -84.

Step 3: Since the slopes of the tangent lines at x = 2 and x = -2 are equal, the tangents are parallel.

Question:

Find the slope of the tangent to the curve y=x^3−3x+2 at the point whose x coordinate is 3.

Answer:

Step 1: Find the derivative of the equation y=x^3−3x+2.

y’ = 3x^2 - 3

Step 2: Substitute x = 3 into the equation y’ = 3x^2 - 3.

y’ = 3(3^2) - 3 y’ = 27 - 3 y’ = 24

Step 3: The slope of the tangent to the curve y=x^3−3x+2 at the point whose x coordinate is 3 is 24.

Question:

The slope of the normal to the curve y=2x^2+3sinx at x=0 is. A 3 B 1​/3 C −3 D −1/3

Answer:

Answer: D −1/3

Step 1: Identify the equation of the curve given in the question.

The equation of the curve is y=2x^2+3sinx.

Step 2: Find the derivative of the equation.

The derivative of the equation is y’=4x+3cosx.

Step 3: Calculate the slope of the normal to the curve at x=0.

The slope of the normal to the curve at x=0 is -1/3.

Therefore, the answer is D −1/3.