Application of Derivatives Exercise 03
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/(x1)^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 pointslope 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 xaxis.
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 xaxis 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:

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

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

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

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

Simplify: y’= (1/2a)t^2

Find the equation of the tangent line: y2at=(1/2a)t^2(xat^2)

Find the equation of the normal line: y2at=(1/2a)t^2(xat^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 xvalues 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:

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

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

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

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 xcoordinate is 2.
Answer:
 Find the equation of the tangent line:
y = 3x^2  1
 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 xaxis 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 xaxis 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 xaxis.
Answer:
 Find the derivative of y = x^3−3x^2−9x+7
dy/dx = 3x^2  6x  9
 Set the derivative equal to 0 in order to find the critical points
3x^2  6x  9 = 0
 Solve the equation to find the critical points
3x^2  6x  9 = 0
(3x + 3)(x  3) = 0
x = 3 and x = 3
 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 xaxis 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:
 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 = (y2y1)/(x2x1)
Therefore, the equation of the line joining the points (2,0) and (4,4) is: y = 2x
 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(x2)
 Set the equation of the line and the equation of the tangent equal to each other and solve for x.
2x = 2(x2) 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 xaxis . (ii) Parallel to yaxis.
Answer:
(i) Parallel to xaxis
The equation of the line parallel to the xaxis 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 xaxis are (1/3, 0) and (1/3, 0).
(ii) Parallel to yaxis
The equation of the line parallel to the yaxis 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 yaxis 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:

Set the equation for the slope of the tangent line equal to the ycoordinate of the point: m = y

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

Set the equation for the slope of the tangent line equal to the equation for the ycoordinate of the point: 3x^2 = x^3

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:
 Set y = 0: 0 = 4x^3 − 2x^5
 Factor the equation: 0 = 2x^5(2−x^2)
 Set each factor equal to zero: 2 = 0 2 − x^2 = 0
 Solve for x: 2 = 0 → x = 0 2 − x^2 = 0 → x = ±√2
 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 pointslope 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/x3 with respect to x.
dy/dx = 1/(x3)^2
Step 2: Set the derivative equal to the slope of the line, which is 2.
1/(x3)^2 = 2
Step 3: Solve for x.
1/(2(x3)^2) = 1
x3 = ±√(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/x3,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:
 Set y=x^3−11x+5 equal to y=x−11
 x^3−12x+16=0
 Factor the equation to get (x4)(x4)(x4)=0
 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:

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

Find the slope of the tangent line. The slope of the tangent line is the derivative of the given equation, which is 2x2.

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.

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:

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.

Now, we can use the slope to find the equation of the tangent line. The equation of a line in slopeintercept form is y=mx+b, where m is the slope and b is the yintercept. 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.

Finally, we can use the slope to find the equation of the normal line. The equation of a line in slopeintercept form is y=mx+b, where m is the slope and b is the yintercept. 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 yintercept.
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 = ((x1)/(x2)x=10+h  (x1)/(x2)x=10)/h
m = ((10+h1)/(10+h2)  (101)/(102))/h
m = (h1)/(h2)
Step 3: Set h = 0.
m = (01)/(02)
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:

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

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

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

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.