Examkin

Tangents and Normals to Curves

24 soru

Tüm türler24Çoktan seçmeli12Sayısal12
Soru 1Soru →

A curve is defined by the equation y=x2−5x+6y = x^2 - 5x + 6. What is the xx-intercept of the line normal to the curve at the point where x=1x = 1?

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Cevap: -5

Cevap

The x-intercept of the normal line to the curve at x = 1 is -5.
Substituting x=1x = 1 into y=x2−5x+6y = x^2 - 5x + 6 yields y=2y = 2, identifying the point (1,2)(1, 2). Differentiating gives dydx=2x−5\frac{dy}{dx} = 2x - 5, which equals −3-3 at x=1x = 1. The normal line gradient is the negative reciprocal, 13\frac{1}{3}. The line equation y−2=13(x−1)y - 2 = \frac{1}{3}(x - 1) simplifies to x−3y+5=0x - 3y + 5 = 0. Setting y=0y = 0 gives x=−5x = -5.

Adım Adım Çözüm

1
Calculate the y-coordinate at x = 1 to determine the point of tangency
Substituting x=1x = 1 into y=x2−5x+6y = x^2 - 5x + 6 gives y=(1)2−5(1)+6=2y = (1)^2 - 5(1) + 6 = 2, yielding the point (1,2)(1, 2).
The normal line intersects the curve at the point of tangency.
2
Find the derivative of the curve and evaluate the tangent slope
dydx=2x−5\frac{dy}{dx} = 2x - 5. At x=1x = 1, mt=2(1)−5=−3m_t = 2(1) - 5 = -3.
The derivative evaluated at a specific point gives the slope of the tangent line to the curve.
3
Compute the slope of the normal line
mn=−1mt=−1−3=13m_n = -\frac{1}{m_t} = -\frac{1}{-3} = \frac{1}{3}.
The normal line is perpendicular to the tangent line, so its gradient is the negative reciprocal of the tangent gradient.
4
Construct the normal line equation and determine its x-intercept
Using point-slope form: y−2=13(x−1)  ⟹  3y−6=x−1  ⟹  x−3y+5=0y - 2 = \frac{1}{3}(x - 1) \implies 3y - 6 = x - 1 \implies x - 3y + 5 = 0. Setting y=0y = 0 gives x+5=0  ⟹  x=−5x + 5 = 0 \implies x = -5.
The xx-intercept occurs where the line crosses the xx-axis, meaning y=0y = 0.

Anahtar Kavram

The slope of the normal line to a curve y=f(x)y = f(x) at x=ax = a is the negative reciprocal of the derivative at that point, mn=−1f′(a)m_n = -\frac{1}{f'(a)}.
Soru 2Soru →

A curve is defined by the equation y=x3−4x+3y = x^3 - 4x + 3. What is the yy-intercept of the tangent line to the curve at the point where x=2x = 2?

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Cevap: -13

Cevap

The y-intercept of the tangent line is −13-13.
At x=2x = 2, substituting into y=x3−4x+3y = x^3 - 4x + 3 yields y=3y = 3. The derivative dydx=3x2−4\frac{dy}{dx} = 3x^2 - 4 evaluated at x=2x = 2 gives m=8m = 8. The tangent line equation is y−3=8(x−2)y - 3 = 8(x - 2), which simplifies to y=8x−13y = 8x - 13. Setting x=0x = 0 gives the yy-intercept of −13-13.

Adım Adım Çözüm

1
Calculate the y-coordinate of the point on the curve at x=2x = 2
y=3y = 3, giving the point (2,3)(2, 3)
The point of contact must lie on the curve.
2
Differentiate the curve equation to find the gradient function
dydx=3x2−4\frac{dy}{dx} = 3x^2 - 4
The derivative of a function gives the gradient of the tangent at any point xx.
3
Evaluate the derivative at x=2x = 2 to find the gradient of the tangent mm
m=8m = 8
Substituting x=2x = 2 into 3x2−43x^2 - 4 yields 3(4)−4=83(4) - 4 = 8.
4
Formulate the linear equation of the tangent line using point (2,3)(2, 3) and gradient m=8m = 8
y=8x−13y = 8x - 13
Applying y−y1=m(x−x1)y - y_1 = m(x - x_1) gives y−3=8(x−2)y - 3 = 8(x - 2), which simplifies to y=8x−13y = 8x - 13.
5
Extract the y-intercept of the tangent line
y-intercept = −13-13
In slope-intercept form y=mx+cy = mx + c, the constant term c=−13c = -13 is the y-intercept.

Anahtar Kavram

Tangents to Curves and Line Intercepts
Soru 3Soru →

Find the gradient of the normal to the curve y=x+1x−1y = \frac{x + 1}{x - 1} at the point where x=3x = 3.

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Cevap: 2

Cevap

The gradient of the normal to the curve at x=3x = 3 is 22.
Differentiating y=x+1x−1y = \frac{x + 1}{x - 1} via the quotient rule gives dydx=−2(x−1)2\frac{dy}{dx} = \frac{-2}{(x - 1)^2}. At x=3x = 3, the tangent gradient is mt=−24=−12m_t = \frac{-2}{4} = -\frac{1}{2}. Since the normal is perpendicular to the tangent, its gradient is mn=−1mt=2m_n = -\frac{1}{m_t} = 2.

Adım Adım Çözüm

1
Differentiate the rational function with respect to xx
Applying the quotient rule ddx(uv)=vu′−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2} where u=x+1u = x + 1 and v=x−1v = x - 1 yields dydx=(x−1)(1)−(x+1)(1)(x−1)2=−2(x−1)2\frac{dy}{dx} = \frac{(x - 1)(1) - (x + 1)(1)}{(x - 1)^2} = \frac{-2}{(x - 1)^2}.
The first derivative determines the slope function of the tangent line to the curve.
2
Substitute x=3x = 3 into the derivative to find the tangent slope mtm_t
mt=−2(3−1)2=−24=−12m_t = \frac{-2}{(3 - 1)^2} = \frac{-2}{4} = -\frac{1}{2}.
Evaluating the derivative at the given xx-coordinate provides the exact gradient of the tangent at that point.
3
Calculate the slope of the normal line mnm_n
mn=−1mt=−1−1/2=2m_n = -\frac{1}{m_t} = -\frac{1}{-1/2} = 2.
The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent slope.

Anahtar Kavram

The slope of the normal line to a curve at a given point is the negative reciprocal of the slope of the tangent line at that point (mn=−1mtm_n = -\frac{1}{m_t}).
Soru 4Soru →

What is the equation of the normal to the curve y=2x2−3x+1y = 2x^2 - 3x + 1 at the point where x=2x = 2?

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Cevap: x+5y−17=0x + 5y - 17 = 0

Cevap

x+5y−17=0x + 5y - 17 = 0
Evaluating y=2x2−3x+1y = 2x^2 - 3x + 1 at x=2x = 2 gives the point (2,3)(2, 3). Differentiating gives dydx=4x−3\frac{dy}{dx} = 4x - 3, which evaluates to 55 at x=2x = 2. The normal line gradient is therefore −15-\frac{1}{5}. Using the point-slope formula y−3=−15(x−2)y - 3 = -\frac{1}{5}(x - 2) and rearranging to standard linear form results in x+5y−17=0x + 5y - 17 = 0.

Adım Adım Çözüm

1
Find the yy-coordinate of the point of contact on the curve when x=2x = 2.
Substitute x=2x = 2 into y=2x2−3x+1y = 2x^2 - 3x + 1: y=2(2)2−3(2)+1=8−6+1=3y = 2(2)^2 - 3(2) + 1 = 8 - 6 + 1 = 3. The point is (2,3)(2, 3).
The line of normal passes through the point of tangency on the curve.
2
Differentiate y=2x2−3x+1y = 2x^2 - 3x + 1 with respect to xx to find the gradient of the tangent.
dydx=4x−3\frac{dy}{dx} = 4x - 3. At x=2x = 2, mt=4(2)−3=5m_t = 4(2) - 3 = 5.
The derivative of a curve evaluated at a point yields the gradient of the tangent line at that point.
3
Determine the gradient of the normal line.
Since the normal is perpendicular to the tangent, mn=−1mt=−15m_n = -\frac{1}{m_t} = -\frac{1}{5}.
Perpendicular lines have gradients whose product is −1-1 (mt⋅mn=−1m_t \cdot m_n = -1).
4
Form the equation of the normal line passing through (2,3)(2, 3) with gradient mn=−15m_n = -\frac{1}{5}.
y−3=−15(x−2)  ⟹  5(y−3)=−(x−2)  ⟹  5y−15=−x+2  ⟹  x+5y−17=0y - 3 = -\frac{1}{5}(x - 2) \implies 5(y - 3) = -(x - 2) \implies 5y - 15 = -x + 2 \implies x + 5y - 17 = 0.
Applying the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) establishes the linear equation.

Anahtar Kavram

Tangents and Normals to Curves
Soru 5Soru →

What is the gradient of the normal to the curve y=x2+2x−1y = x^2 + 2x - 1 at the point where x=1x = 1?

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Cevap: −14-\frac{1}{4}

Cevap

The gradient of the normal to the curve at x=1x = 1 is −14-\frac{1}{4}.
Differentiating y=x2+2x−1y = x^2 + 2x - 1 gives dydx=2x+2\frac{dy}{dx} = 2x + 2. Substituting x=1x = 1 yields a tangent slope of 44. Taking the negative reciprocal gives −14-\frac{1}{4}, which is the correct gradient of the normal.

Adım Adım Çözüm

1
Find the derivative of the curve equation to determine the general tangent gradient function.
dydx=2x+2\frac{dy}{dx} = 2x + 2
The derivative of a function gives the slope of the tangent line at any point xx.
2
Evaluate the tangent gradient at the specific point x=1x = 1.
m=2(1)+2=4m = 2(1) + 2 = 4
Substituting x=1x = 1 into the derivative gives the slope of the tangent at that specific x-coordinate.
3
Calculate the gradient of the normal line using the perpendicular condition.
mn=−1m=−14m_n = -\frac{1}{m} = -\frac{1}{4}
The normal line is perpendicular to the tangent line, so its gradient is the negative reciprocal of the tangent gradient.

Anahtar Kavram

Relationship between tangent and normal gradients (mnormal=−1mtangentm_{\text{normal}} = -\frac{1}{m_{\text{tangent}}})
Tahmini Süre:1m 0s
Soru 6Soru →

What is the yy-intercept of the normal line to the curve y=2x−1x+1y = \frac{2x - 1}{x + 1} at the point where the curve crosses the xx-axis?

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Cevap: 38\frac{3}{8}

Cevap

The yy-intercept of the normal line is 38\frac{3}{8}.
To find the yy-intercept of the normal line, set y=0y = 0 to find the point of contact on the xx-axis, which gives (12,0)\left(\frac{1}{2}, 0\right). Differentiating y=2x−1x+1y = \frac{2x - 1}{x + 1} gives dydx=3(x+1)2\frac{dy}{dx} = \frac{3}{(x + 1)^2}. At x=12x = \frac{1}{2}, the tangent slope is 43\frac{4}{3}, making the normal slope −34-\frac{3}{4}. The equation of the normal line is y=−34x+38y = -\frac{3}{4}x + \frac{3}{8}, so its yy-intercept is 38\frac{3}{8}.

Adım Adım Çözüm

1
Find the point of intersection of the curve with the xx-axis
Set y=0  ⟹  2x−1x+1=0  ⟹  2x−1=0  ⟹  x=12y = 0 \implies \frac{2x - 1}{x + 1} = 0 \implies 2x - 1 = 0 \implies x = \frac{1}{2}. The point is (12,0)\left(\frac{1}{2}, 0\right).
The normal line is drawn at the point where the curve crosses the xx-axis.
2
Differentiate y=2x−1x+1y = \frac{2x - 1}{x + 1} using the quotient rule
\(\frac{dy}{dx} = \frac{2(x + 1) - (2x - 1)(1)}{(x + 1)^2} = \frac{2x + 2 - 2x + 1}{(x + 1)^2} = \frac{3}{(x + 1)^2}\)
The derivative provides the gradient function of the tangent line to the curve.
3
Evaluate the gradient of the tangent and normal lines at x=12x = \frac{1}{2}
Tangent gradient mt=3(12+1)2=394=43m_t = \frac{3}{\left(\frac{1}{2} + 1\right)^2} = \frac{3}{\frac{9}{4}} = \frac{4}{3}. Normal gradient mn=−1mt=−34m_n = -\frac{1}{m_t} = -\frac{3}{4}.
The normal line is perpendicular to the tangent line at the point of contact.
4
Determine the equation of the normal line and evaluate its yy-intercept
Using y−0=−34(x−12)  ⟹  y=−34x+38y - 0 = -\frac{3}{4}\left(x - \frac{1}{2}\right) \implies y = -\frac{3}{4}x + \frac{3}{8}. Setting x=0x = 0 gives y=38y = \frac{3}{8}.
The yy-intercept is the value of yy when x=0x = 0 on the line.

Anahtar Kavram

Tangents and Normals to Curves
Soru 7Soru →

What is the equation of the normal line to the curve y=x3−3x+2y = x^3 - 3x + 2 at the point where x=2x = 2?

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Cevap: x+9y−38=0x + 9y - 38 = 0

Cevap

The equation of the normal line is x+9y−38=0x + 9y - 38 = 0.
Evaluating yy at x=2x = 2 gives the point (2,4)(2, 4). Differentiating y=x3−3x+2y = x^3 - 3x + 2 yields dydx=3x2−3\frac{dy}{dx} = 3x^2 - 3. At x=2x = 2, the tangent slope is mt=9m_t = 9, making the normal slope mn=−1/9m_n = -1/9. Using the point-slope formula y−4=−1/9(x−2)y - 4 = -1/9(x - 2) simplifies to x+9y−38=0x + 9y - 38 = 0.

Adım Adım Çözüm

1
Find the yy-coordinate at x=2x = 2
y=(2)3−3(2)+2=8−6+2=4y = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4. The point on the curve is (2,4)(2, 4).
The point of tangency/normal line intersection must be determined on the curve.
2
Calculate the derivative dydx\frac{dy}{dx} to find the slope of the tangent line
dydx=3x2−3\frac{dy}{dx} = 3x^2 - 3. At x=2x = 2, dydx=3(2)2−3=12−3=9\frac{dy}{dx} = 3(2)^2 - 3 = 12 - 3 = 9.
The derivative evaluated at x=2x = 2 gives the gradient of the tangent line.
3
Find the slope of the normal line
mn=−1mt=−19m_n = -\frac{1}{m_t} = -\frac{1}{9}.
The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent slope.
4
Use the point-slope form to find the equation of the normal line
y−4=−19(x−2)  ⟹  9(y−4)=−(x−2)  ⟹  9y−36=−x+2  ⟹  x+9y−38=0y - 4 = -\frac{1}{9}(x - 2) \implies 9(y - 4) = -(x - 2) \implies 9y - 36 = -x + 2 \implies x + 9y - 38 = 0.
Substituting point (2,4)(2, 4) and slope −1/9-1/9 into y−y1=mn(x−x1)y - y_1 = m_n(x - x_1) yields the standard line equation.

Anahtar Kavram

The slope of the normal line to a curve y=f(x)y = f(x) at (x1,y1)(x_1, y_1) is mn=−1f′(x1)m_n = -\frac{1}{f'(x_1)}.
Soru 8Soru →

A curve has the equation y=13x3−2x2+3x+1y = \frac{1}{3}x^3 - 2x^2 + 3x + 1. What is the positive xx-coordinate of the point on the curve where the tangent line is parallel to the line y=8x−5y = 8x - 5?

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Cevap: 5

Cevap

The positive xx-coordinate of the point where the tangent is parallel to the line is 55.
The gradient of the line y=8x−5y = 8x - 5 is 88. Differentiating y=13x3−2x2+3x+1y = \frac{1}{3}x^3 - 2x^2 + 3x + 1 gives dydx=x2−4x+3\frac{dy}{dx} = x^2 - 4x + 3. Setting dydx=8\frac{dy}{dx} = 8 leads to x2−4x−5=0x^2 - 4x - 5 = 0, which factors as (x−5)(x+1)=0(x - 5)(x + 1) = 0. The solutions are x=5x = 5 and x=−1x = -1. Selecting the positive value gives x=5x = 5.

Adım Adım Çözüm

1
Find the gradient (slope) of the given straight line.
The line equation is in slope-intercept form y=mx+cy = mx + c, where the slope m=8m = 8.
Parallel lines have equal slopes, so the gradient of the tangent to the curve must equal 8.
2
Differentiate the curve equation to find the gradient function dydx\frac{dy}{dx}.
dydx=ddx(13x3−2x2+3x+1)=x2−4x+3\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{3}x^3 - 2x^2 + 3x + 1\right) = x^2 - 4x + 3.
The first derivative of a curve represents the gradient of the tangent at any point xx.
3
Equate the derivative to the slope of the line and solve for xx.
x2−4x+3=8  ⟹  x2−4x−5=0  ⟹  (x−5)(x+1)=0x^2 - 4x + 3 = 8 \implies x^2 - 4x - 5 = 0 \implies (x - 5)(x + 1) = 0. The roots are x=5x = 5 and x=−1x = -1.
Solving the quadratic equation gives all xx-values where the tangent line has a slope of 8.
4
Select the positive xx-coordinate as requested.
x=5x = 5.
The question specifically asks for the positive value among the solutions.

Anahtar Kavram

Finding points on a curve where the tangent is parallel to a given line
Soru 9Soru →

The line y=5x+cy = 5x + c is tangent to the curve y=x3−7x+15y = x^3 - 7x + 15 at a point PP located in the first quadrant. What is the equation of the normal line to the curve at point PP?

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Cevap: x+5y−47=0x + 5y - 47 = 0

Cevap

The equation of the normal line to the curve at point P is x+5y−47=0x + 5y - 47 = 0.
Differentiating the curve yields dydx=3x2−7\frac{dy}{dx} = 3x^2 - 7. Equating this derivative to the tangent slope of 5 gives x=2x = 2 for the first quadrant. Evaluating the curve equation at x=2x = 2 gives y=9y = 9, locating point P(2,9)P(2, 9). The normal gradient is the negative reciprocal of the tangent slope, giving mn=−15m_n = -\frac{1}{5}. Applying the point-slope formula with (2,9)(2, 9) yields y−9=−15(x−2)y - 9 = -\frac{1}{5}(x - 2), which simplifies to x+5y−47=0x + 5y - 47 = 0.

Adım Adım Çözüm

1
Find the x-coordinate of the point of tangency P
x=2x = 2
The gradient of the curve dydx=3x2−7\frac{dy}{dx} = 3x^2 - 7 must equal the gradient of the given line m=5m = 5. Solving 3x2−7=53x^2 - 7 = 5 gives 3x2=12  ⟹  x2=4  ⟹  x=23x^2 = 12 \implies x^2 = 4 \implies x = 2 (since PP is in the first quadrant).
2
Calculate the y-coordinate of point P
y=9y = 9
Substitute x=2x = 2 into the curve equation y=23−7(2)+15=8−14+15=9y = 2^3 - 7(2) + 15 = 8 - 14 + 15 = 9. Thus, P=(2,9)P = (2, 9).
3
Determine the gradient of the normal line
mn=−15m_n = -\frac{1}{5}
The normal line is perpendicular to the tangent line. Since the tangent gradient mt=5m_t = 5, the normal gradient is mn=−1mt=−15m_n = -\frac{1}{m_t} = -\frac{1}{5}.
4
Formulate the equation of the normal line
x+5y−47=0x + 5y - 47 = 0
Using the point-slope equation y−y1=mn(x−x1)y - y_1 = m_n(x - x_1) with point (2,9)(2, 9) and gradient −15-\frac{1}{5}: y−9=−15(x−2)  ⟹  5(y−9)=−(x−2)  ⟹  5y−45=−x+2  ⟹  x+5y−47=0y - 9 = -\frac{1}{5}(x - 2) \implies 5(y - 9) = -(x - 2) \implies 5y - 45 = -x + 2 \implies x + 5y - 47 = 0.

Anahtar Kavram

The gradient of the normal to a curve at a given point is the negative reciprocal of the derivative (tangent gradient) at that point: mnormal=−1dydxm_{normal} = -\frac{1}{\frac{dy}{dx}}.
Soru 10Soru →

What is the xx-intercept of the normal line to the curve y=x2−4x+5y = x^2 - 4x + 5 at the point where x=3x = 3?

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Cevap: 7

Cevap

The xx-intercept of the normal line is 7.
Differentiating y=x2−4x+5y = x^2 - 4x + 5 gives dydx=2x−4\frac{dy}{dx} = 2x - 4. Substituting x=3x = 3 yields a tangent gradient of 22, so the normal gradient is −12-\frac{1}{2}. The curve passes through (3,2)(3, 2) at x=3x = 3. The equation of the normal is y−2=−12(x−3)y - 2 = -\frac{1}{2}(x - 3), which simplifies to x+2y−7=0x + 2y - 7 = 0. Setting y=0y = 0 gives x=7x = 7.

Adım Adım Çözüm

1
Find the yy-coordinate corresponding to x=3x = 3
y=32−4(3)+5=9−12+5=2y = 3^2 - 4(3) + 5 = 9 - 12 + 5 = 2. The point on the curve is (3,2)(3, 2).
The normal line passes through the specific point of tangency on the curve.
2
Calculate the gradient of the tangent and normal lines at x=3x = 3
dydx=2x−4\frac{dy}{dx} = 2x - 4. At x=3x = 3, mt=2(3)−4=2m_t = 2(3) - 4 = 2. Therefore, mn=−1mt=−12m_n = -\frac{1}{m_t} = -\frac{1}{2}.
The normal line is perpendicular to the tangent line.
3
Determine the equation of the normal line
y−2=−12(x−3)  ⟹  2(y−2)=−(x−3)  ⟹  x+2y−7=0y - 2 = -\frac{1}{2}(x - 3) \implies 2(y - 2) = -(x - 3) \implies x + 2y - 7 = 0.
Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) to write the linear equation.
4
Find the xx-intercept of the normal line
Set y=0y = 0: x+2(0)−7=0  ⟹  x=7x + 2(0) - 7 = 0 \implies x = 7.
The xx-intercept occurs where y=0y = 0.

Anahtar Kavram

Tangents and Normals to Curves
Tahmini Süre:1m 30s
Soru 11Soru →

Find the yy-intercept of the tangent line to the curve y=x3−6x+8y = x^3 - 6x + 8 at the point where x=1x = 1.

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Cevap: 6

Cevap

The yy-intercept of the tangent line is 6.
Differentiating y=x3−6x+8y = x^3 - 6x + 8 yields dydx=3x2−6\frac{dy}{dx} = 3x^2 - 6. At x=1x = 1, the point on the curve is (1,3)(1, 3) and the gradient of the tangent is m=−3m = -3. Substituting into the point-slope form y−3=−3(x−1)y - 3 = -3(x - 1) gives y=−3x+6y = -3x + 6. Setting x=0x = 0 identifies the yy-intercept as 6.

Adım Adım Çözüm

1
Find the point of tangency on the curve
For x=1x = 1, y=(1)3−6(1)+8=3y = (1)^3 - 6(1) + 8 = 3. The point is (1,3)(1, 3).
The tangent line touches the curve at the point corresponding to x=1x = 1.
2
Find the gradient function of the curve
dydx=3x2−6\frac{dy}{dx} = 3x^2 - 6
The derivative of the curve equation gives the gradient of the tangent at any point.
3
Calculate the gradient of the tangent line at x=1x = 1
m = 3(1)^2 - 6 = -3
Substitute x=1x = 1 into the derivative.
4
Formulate the equation of the tangent line
y - 3 = -3(x - 1) \implies y = -3x + 6
Apply the point-slope line equation y−y1=m(x−x1)y - y_1 = m(x - x_1) using (1,3)(1, 3) and m=−3m = -3.
5
Determine the yy-intercept
Setting x=0x = 0 yields y=6y = 6.
The yy-intercept is the value of yy when x=0x = 0.

Anahtar Kavram

Tangents to Curves and Axis Intercepts
Soru 12Soru →

A normal line is drawn to the curve y=x2−5x−3y = \frac{x^2 - 5}{x - 3} at the point where x=2x = 2. What is the xx-intercept of this normal line?

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Cevap: -1

Cevap

The xx-intercept of the normal line is −1-1.
Substituting x=2x = 2 into the curve yields the point (2,1)(2, 1). Differentiation via the quotient rule gives dydx=x2−6x+5(x−3)2\frac{dy}{dx} = \frac{x^2 - 6x + 5}{(x - 3)^2}. At x=2x = 2, the tangent slope is −3-3, making the normal slope 13\frac{1}{3}. The line equation y−1=13(x−2)y - 1 = \frac{1}{3}(x - 2) simplifies to x−3y+1=0x - 3y + 1 = 0. Setting y=0y = 0 gives x=−1x = -1.

Adım Adım Çözüm

1
Calculate the y-coordinate of the point of tangency
y=1y = 1
Substitute x=2x = 2 into the curve equation y=x2−5x−3y = \frac{x^2 - 5}{x - 3} to get the point (2,1)(2, 1).
2
Differentiate the function using the quotient rule
dydx=x2−6x+5(x−3)2\frac{dy}{dx} = \frac{x^2 - 6x + 5}{(x - 3)^2}
Applying ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} to u=x2−5u = x^2 - 5 and v=x−3v = x - 3.
3
Find the tangent gradient at x=2x = 2
mt=−3m_t = -3
Evaluate dydx\frac{dy}{dx} at x=2x = 2 to obtain mt=22−6(2)+5(2−3)2=−3m_t = \frac{2^2 - 6(2) + 5}{(2-3)^2} = -3.
4
Calculate the gradient of the normal line
mn=13m_n = \frac{1}{3}
The normal line is perpendicular to the tangent line, so mn=−1mt=13m_n = -\frac{1}{m_t} = \frac{1}{3}.
5
Derive the line equation for the normal
x−3y+1=0x - 3y + 1 = 0
Use point-slope form y−y1=mn(x−x1)y - y_1 = m_n(x - x_1) with (2,1)(2, 1) and mn=13m_n = \frac{1}{3}.
6
Find the x-intercept
x=−1x = -1
Set y=0y = 0 in the normal equation x−3(0)+1=0x - 3(0) + 1 = 0, giving x=−1x = -1.

Anahtar Kavram

Equation of normal line to a rational curve and finding its axis intercepts
Soru 13Soru →

What is the gradient of the tangent line to the curve y=4x2−7x+2y = 4x^2 - 7x + 2 at the point where x=3x = 3?

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Cevap: 17

Cevap

The gradient of the tangent line to the curve at x=3x = 3 is 17.
Differentiating y=4x2−7x+2y = 4x^2 - 7x + 2 gives dydx=8x−7\frac{dy}{dx} = 8x - 7. Substituting x=3x = 3 yields 8(3)−7=178(3) - 7 = 17.

Adım Adım Çözüm

1
Find the derivative of the curve's equation
dydx=8x−7\frac{dy}{dx} = 8x - 7
The gradient of the curve at any point is given by its first derivative with respect to x.
2
Evaluate the derivative at x = 3
m = 8(3) - 7 = 17
Substituting the given x-coordinate into the derivative gives the specific slope of the tangent line at that point.

Anahtar Kavram

The gradient of the tangent to a curve y=f(x)y = f(x) at x=ax = a is the value of the first derivative f′(a)f'(a).
Soru 14Soru →

Find the xx-intercept of the normal line to the curve y=x3−3x2+4x−1y = x^3 - 3x^2 + 4x - 1 at the point where x=1x = 1.

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Cevap: 2

Cevap

The xx-intercept of the normal line is 22.
At x=1x = 1, the point on the curve y=x3−3x2+4x−1y = x^3 - 3x^2 + 4x - 1 is (1,1)(1, 1). The derivative is dydx=3x2−6x+4\frac{dy}{dx} = 3x^2 - 6x + 4, which equals 11 at x=1x = 1. The normal gradient is therefore −1-1. The normal line equation is y−1=−1(x−1)y - 1 = -1(x - 1), which simplifies to y=−x+2y = -x + 2. Setting y=0y = 0 gives x=2x = 2.

Adım Adım Çözüm

1
Find the y-coordinate at x = 1
y = 1, so the point on the curve is (1, 1)
The point of contact is needed to construct the equation of the normal line.
2
Differentiate the function with respect to x
dy/dx = 3x^2 - 6x + 4
The derivative gives the expression for the gradient of the tangent to the curve.
3
Find the gradient of the tangent and normal at x = 1
Gradient of tangent m_t = 1; gradient of normal m_n = -1
The normal line is perpendicular to the tangent line, so m_n = -1 / m_t.
4
Determine the equation of the normal line
y - 1 = -1(x - 1) => y = -x + 2
Applying the straight-line equation y - y_1 = m(x - x_1).
5
Find the x-intercept of the normal line
x = 2
Setting y = 0 in the normal line equation yields the x-intercept.

Anahtar Kavram

Equation and axis intercepts of a normal line to a curve
Soru 15Soru →

What is the gradient of the normal to the curve y=x2−4x+5y = x^2 - 4x + 5 at the point where x=3x = 3?

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Cevap: −12-\frac{1}{2}

Cevap

The gradient of the normal to the curve is −12-\frac{1}{2}.
Differentiating y=x2−4x+5y = x^2 - 4x + 5 gives dydx=2x−4\frac{dy}{dx} = 2x - 4. Substituting x=3x = 3 yields a tangent gradient of 22. Since the normal line is perpendicular to the tangent line, its gradient is the negative reciprocal, −12-\frac{1}{2}.

Adım Adım Çözüm

1
Differentiate the equation of the curve to determine the gradient function.
dydx=2x−4\frac{dy}{dx} = 2x - 4
The first derivative represents the gradient of the tangent to the curve at any point xx.
2
Evaluate the derivative at x=3x = 3 to find the tangent gradient mtm_t.
mt=2(3)−4=2m_t = 2(3) - 4 = 2
Substituting x=3x = 3 gives the slope of the tangent line at the given point.
3
Calculate the gradient of the normal mnm_n using mn=−1mtm_n = -\frac{1}{m_t}.
mn=−12m_n = -\frac{1}{2}
The normal line is perpendicular to the tangent line, so its gradient is the negative reciprocal of the tangent's gradient.

Anahtar Kavram

Gradient of a Normal Line
Soru 16Soru →

At what point on the curve y=2x2−8x+5y = 2x^2 - 8x + 5 is the normal line parallel to the straight line x+4y−7=0x + 4y - 7 = 0?

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Cevap: (3,−1)(3, -1)

Cevap

The point of contact on the curve is (3,−1)(3, -1).
The line x+4y−7=0x + 4y - 7 = 0 has a slope of −14-\frac{1}{4}. Since the normal line is parallel to this line, the normal gradient is mn=−14m_n = -\frac{1}{4}. Consequently, the tangent gradient must be mt=4m_t = 4 because mt⋅mn=−1m_t \cdot m_n = -1. Equating the derivative dydx=4x−8\frac{dy}{dx} = 4x - 8 to 44 yields x=3x = 3. Substituting x=3x = 3 into the curve equation y=2x2−8x+5y = 2x^2 - 8x + 5 gives y=−1y = -1, yielding the point (3,−1)(3, -1).

Adım Adım Çözüm

1
Find the gradient of the given straight line.
Rearranging x+4y−7=0x + 4y - 7 = 0 into slope-intercept form gives y=−14x+74y = -\frac{1}{4}x + \frac{7}{4}, so the line's gradient is m=−14m = -\frac{1}{4}.
Parallel lines have equal gradients, so the normal line to the curve must have gradient mn=−14m_n = -\frac{1}{4}.
2
Determine the required gradient of the tangent line.
Since mt⋅mn=−1m_t \cdot m_n = -1, we have mt=−1−14=4m_t = -\frac{1}{-\frac{1}{4}} = 4.
The tangent and normal lines are perpendicular to each other.
3
Differentiate the curve's equation to find the xx-coordinate.
dydx=4x−8\frac{dy}{dx} = 4x - 8. Setting dydx=4\frac{dy}{dx} = 4 gives 4x−8=4  ⟹  4x=12  ⟹  x=34x - 8 = 4 \implies 4x = 12 \implies x = 3.
The derivative represents the slope of the tangent line at any point xx.
4
Substitute x=3x = 3 back into the curve's equation to find yy.
y=2(3)2−8(3)+5=2(9)−24+5=18−24+5=−1y = 2(3)^2 - 8(3) + 5 = 2(9) - 24 + 5 = 18 - 24 + 5 = -1.
The point of contact lies on the original curve.

Anahtar Kavram

Relationship between gradients of parallel lines, tangent lines, and normal lines to a curve
Soru 17Soru →

What is the yy-intercept of the tangent line to the curve y=x3−2x+4y = x^3 - 2x + 4 at the point where x=1x = 1?

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Cevap: 2

Cevap

The y-intercept of the tangent line is 2.
The curve evaluated at x=1x=1 gives point (1,3)(1,3). The derivative y′=3x2−2y'=3x^2-2 gives slope m=1m=1 at x=1x=1. The tangent line equation is y−3=1(x−1)y-3=1(x-1), which simplifies to y=x+2y=x+2. The yy-intercept occurs at x=0x=0, giving y=2y=2.

Adım Adım Çözüm

1
Find the y-coordinate of the point on the curve at x=1x = 1.
At x=1x = 1, y=(1)3−2(1)+4=3y = (1)^3 - 2(1) + 4 = 3. The point of tangency is (1,3)(1, 3).
The point of tangency lies on the curve.
2
Find the gradient function of the curve by differentiation.
dydx=3x2−2\frac{dy}{dx} = 3x^2 - 2.
The first derivative represents the slope of the tangent line.
3
Calculate the slope of the tangent line at x=1x = 1.
m=3(1)2−2=1m = 3(1)^2 - 2 = 1.
Substituting x=1x = 1 into the derivative yields the slope at that specific point.
4
Formulate the equation of the tangent line.
y−3=1(x−1)  ⟹  y=x+2y - 3 = 1(x - 1) \implies y = x + 2.
Use point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with point (1,3)(1,3) and slope m=1m=1.
5
Find the yy-intercept of the tangent line.
Setting x=0x = 0 gives y=2y = 2.
The yy-intercept is the value of yy where the line crosses the vertical axis.

Anahtar Kavram

Tangents and Normals to Curves
Soru 18Soru →

What is the gradient of the normal line to the curve y=x2−3x+5y = x^2 - 3x + 5 at the point where x=1x = 1?

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Cevap: 1

Cevap

The gradient of the normal line to the curve at x=1x = 1 is 1.
Differentiating y=x2−3x+5y = x^2 - 3x + 5 gives dydx=2x−3\frac{dy}{dx} = 2x - 3. Evaluating this derivative at x=1x = 1 gives the tangent gradient mt=−1m_t = -1. Because the normal line is perpendicular to the tangent, its gradient is mn=−1mt=−1−1=1m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1.

Adım Adım Çözüm

1
Differentiate the function with respect to x
dydx=2x−3\frac{dy}{dx} = 2x - 3
The derivative of a function gives the slope of the tangent line at any given x-coordinate.
2
Evaluate the derivative at x = 1
m_t = -1
Substituting the given point's x-coordinate into the gradient function yields the slope of the tangent.
3
Calculate the negative reciprocal of the tangent slope
m_n = 1
Since the normal line is perpendicular to the tangent line, its gradient is m_n = -1 / m_t.

Anahtar Kavram

Gradient of a Normal Line
Soru 19Soru →

The line 3x−y+k=03x - y + k = 0 is a normal to the curve y=x+2x−1y = \frac{x + 2}{x - 1} at a point PP located in the first quadrant. What is the value of the constant kk?

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Cevap: -10

Cevap

The value of the constant kk is −10-10.
Differentiating y=x+2x−1y = \frac{x + 2}{x - 1} yields dydx=−3(x−1)2\frac{dy}{dx} = \frac{-3}{(x - 1)^2}. The gradient of the normal line is mN=−1dy/dx=(x−1)23m_N = -\frac{1}{dy/dx} = \frac{(x - 1)^2}{3}. Equating mNm_N to the gradient of 3x−y+k=03x - y + k = 0 (which is 33) gives (x−1)23=3  ⟹  (x−1)2=9\frac{(x - 1)^2}{3} = 3 \implies (x - 1)^2 = 9. Solving gives x=4x = 4 or x=−2x = -2. Since point PP lies in the first quadrant, x=4x = 4, which gives y=4+24−1=2y = \frac{4 + 2}{4 - 1} = 2. Substituting (4,2)(4, 2) into 3x−y+k=03x - y + k = 0 gives 3(4)−2+k=03(4) - 2 + k = 0, so k=−10k = -10.

Adım Adım Çözüm

1
Differentiate the curve equation to find the tangent gradient function.
Using the quotient rule on y=x+2x−1y = \frac{x + 2}{x - 1}, dydx=1(x−1)−1(x+2)(x−1)2=−3(x−1)2\frac{dy}{dx} = \frac{1(x - 1) - 1(x + 2)}{(x - 1)^2} = \frac{-3}{(x - 1)^2}.
The derivative gives the gradient of the tangent line to the curve at any point xx.
2
Determine the gradient function for the normal line.
The normal gradient is mN=−1dy/dx=(x−1)23m_N = -\frac{1}{dy/dx} = \frac{(x - 1)^2}{3}.
The normal line is perpendicular to the tangent line, so its gradient is the negative reciprocal of dydx\frac{dy}{dx}.
3
Equate the normal gradient to the gradient of the given line and solve for xx.
The line 3x−y+k=03x - y + k = 0 has gradient 33. Setting (x−1)23=3  ⟹  (x−1)2=9  ⟹  x−1=±3\frac{(x - 1)^2}{3} = 3 \implies (x - 1)^2 = 9 \implies x - 1 = \pm 3, giving x=4x = 4 or x=−2x = -2.
The normal line at point PP must be parallel to (and thus have the same gradient as) the given line 3x−y+k=03x - y + k = 0.
4
Select the coordinate corresponding to the first quadrant and calculate the yy-coordinate.
Since PP lies in the first quadrant (x>0,y>0x > 0, y > 0), x=4x = 4. Substituting x=4x = 4 into the curve equation gives y=4+24−1=2y = \frac{4 + 2}{4 - 1} = 2. Thus, P=(4,2)P = (4, 2).
The problem specifies that point PP is located in the first quadrant.
5
Substitute point P(4,2)P(4, 2) into the line equation to solve for kk.
Substituting x=4x = 4 and y=2y = 2 into 3x−y+k=03x - y + k = 0 yields 3(4)−2+k=0  ⟹  10+k=0  ⟹  k=−103(4) - 2 + k = 0 \implies 10 + k = 0 \implies k = -10.
Point PP lies on the normal line, so its coordinates must satisfy the line equation.

Anahtar Kavram

The normal line to a curve at a point PP is perpendicular to the tangent line at PP, with gradient mnormal=−1dy/dxm_{\text{normal}} = -\frac{1}{dy/dx}. A point PP on the curve must satisfy both the curve equation and the normal line equation.
Soru 20Soru →

A normal line is drawn to the curve y=x2+3x−1y = \frac{x^2 + 3}{x - 1} at the point where x=2x = 2. Calculate the xx-intercept of this normal line.

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Cevap: -19

Cevap

The x-intercept of the normal line is -19.
Substituting x=2x = 2 into the curve function gives y=7y = 7. Differentiating via the quotient rule yields dydx=x2−2x−3(x−1)2\frac{dy}{dx} = \frac{x^2 - 2x - 3}{(x - 1)^2}, which equals −3-3 at x=2x = 2. The perpendicular slope of the normal line is 13\frac{1}{3}. Constructing the line equation through (2,7)(2, 7) gives x−3y+19=0x - 3y + 19 = 0. Setting y=0y = 0 produces x=−19x = -19.

Adım Adım Çözüm

1
Find the point of contact by substituting x=2x = 2 into the curve equation
y=22+32−1=71=7y = \frac{2^2 + 3}{2 - 1} = \frac{7}{1} = 7, so the point is (2,7)(2, 7)
The line is drawn at x=2x = 2, so we need the full coordinate pair (x1,y1)(x_1, y_1)
2
Differentiate the curve y=x2+3x−1y = \frac{x^2 + 3}{x - 1} using the quotient rule
\frac{dy}{dx} = \frac{(x - 1)(2x) - (x^2 + 3)(1)}{(x - 1)^2} = \frac{x^2 - 2x - 3}{(x - 1)^2}
The derivative gives the gradient function of the curve
3
Evaluate the tangent slope mtm_t at x=2x = 2
m_t = \frac{2^2 - 2(2) - 3}{(2 - 1)^2} = \frac{-3}{1} = -3
Evaluating the derivative yields the gradient of the tangent at the point
4
Determine the slope of the normal line mnm_n
m_n = -\frac{1}{m_t} = -\frac{1}{-3} = \frac{1}{3}
Normal lines are perpendicular to tangent lines, so mn⋅mt=−1m_n \cdot m_t = -1
5
Formulate the equation of the normal line passing through (2,7)(2, 7)
y - 7 = \frac{1}{3}(x - 2) \implies 3y - 21 = x - 2 \implies x - 3y + 19 = 0
Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1)
6
Find the xx-intercept by setting y=0y = 0
x - 3(0) + 19 = 0 \implies x = -19
The xx-intercept occurs where the line crosses the xx-axis (y=0y = 0)

Anahtar Kavram

Finding the equation and x-intercept of a normal line to a curve
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Tangents and Normals to Curves Alıştırma Soruları — JAMB UTME | Examkin