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Tangents and Normals to Curves

24 soru

Tüm türler24Çoktan seçmeli12Sayısal12
Soru 21Soru →

What are the coordinates of the point on the curve y=2x2−5x+1y = 2x^2 - 5x + 1 where the tangent line is perpendicular to the line x+3y−4=0x + 3y - 4 = 0?

Cevabı ve açıklamayı göster

Cevap: (2,−1)(2, -1)

Cevap

The point on the curve is (2,−1)(2, -1).
Rearranging the line equation x+3y−4=0x + 3y - 4 = 0 gives a gradient of −13-\frac{1}{3}. The tangent line is perpendicular, so its gradient must be 33. Differentiating y=2x2−5x+1y = 2x^2 - 5x + 1 gives dydx=4x−5\frac{dy}{dx} = 4x - 5. Setting 4x−5=34x - 5 = 3 gives x=2x = 2. Substituting x=2x = 2 into the curve equation yields y=2(2)2−5(2)+1=−1y = 2(2)^2 - 5(2) + 1 = -1. Thus, the point is (2,−1)(2, -1).

Adım Adım Çözüm

1
Determine the gradient of the given line.
Rearranging x+3y−4=0x + 3y - 4 = 0 into slope-intercept form gives y=−13x+43y = -\frac{1}{3}x + \frac{4}{3}, so the gradient is m1=−13m_1 = -\frac{1}{3}.
The slope of a linear equation Ax+By+C=0Ax + By + C = 0 is −AB-\frac{A}{B}.
2
Calculate the gradient of the tangent line.
Since the tangent line is perpendicular to the given line, its gradient is mT=−1m1=3m_T = -\frac{1}{m_1} = 3.
Perpendicular lines have gradients whose product is −1-1 (m1⋅m2=−1m_1 \cdot m_2 = -1).
3
Find the derivative of the curve and set it equal to the tangent gradient.
dydx=ddx(2x2−5x+1)=4x−5\frac{dy}{dx} = \frac{d}{dx}(2x^2 - 5x + 1) = 4x - 5. Setting 4x−5=34x - 5 = 3 yields 4x=8  ⟹  x=24x = 8 \implies x = 2.
The derivative dydx\frac{dy}{dx} gives the gradient of the tangent to the curve at any point xx.
4
Substitute the xx-coordinate into the original curve equation to find yy.
y=2(2)2−5(2)+1=8−10+1=−1y = 2(2)^2 - 5(2) + 1 = 8 - 10 + 1 = -1.
The point lies on the curve, so its coordinates must satisfy the curve's equation.

Anahtar Kavram

Tangents and Normals to Curves
Soru 22Soru →

What is the yy-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?

Cevabı ve açıklamayı göster

Cevap: 22

Cevap

The yy-intercept of the normal line is 22.
Evaluating y=x3−3x2+4x−1y = x^3 - 3x^2 + 4x - 1 at x=1x = 1 yields y=1y = 1. Differentiating gives dydx=3x2−6x+4\frac{dy}{dx} = 3x^2 - 6x + 4, which evaluates to 11 at x=1x = 1. Since the normal is perpendicular to the tangent, its gradient is −1-1. Substituting into y−1=−1(x−1)y - 1 = -1(x - 1) yields y=−x+2y = -x + 2, giving a yy-intercept of 22.

Adım Adım Çözüm

1
Find the yy-coordinate of the point of tangency.
At x=1x = 1, y=(1)3−3(1)2+4(1)−1=1−3+4−1=1y = (1)^3 - 3(1)^2 + 4(1) - 1 = 1 - 3 + 4 - 1 = 1. Point of contact is (1,1)(1, 1).
The point must lie on the curve.
2
Differentiate yy with respect to xx to find the gradient function.
\frac{dy}{dx} = 3x^2 - 6x + 4.
The first derivative represents the gradient of the tangent to the curve.
3
Calculate the gradient of the tangent and normal at x=1x = 1.
Tangent gradient mt=3(1)2−6(1)+4=1m_t = 3(1)^2 - 6(1) + 4 = 1. Normal gradient mn=−1mt=−1m_n = -\frac{1}{m_t} = -1.
The normal line is perpendicular to the tangent line, so mn⋅mt=−1m_n \cdot m_t = -1.
4
Determine the equation of the normal line and find its yy-intercept.
Using y−y1=mn(x−x1)  ⟹  y−1=−1(x−1)  ⟹  y=−x+2y - y_1 = m_n(x - x_1) \implies y - 1 = -1(x - 1) \implies y = -x + 2. Setting x=0x = 0 gives y=2y = 2.
The yy-intercept occurs where the line crosses the yy-axis (x=0x = 0).

Anahtar Kavram

Tangents and Normals to Curves
Soru 23Soru →

Find the equation of the normal to the curve y=2sin⁡x−cos⁡xy = 2\sin x - \cos x at the point where x=0x = 0.

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Cevap: x+2y+2=0x + 2y + 2 = 0

Cevap

x+2y+2=0x + 2y + 2 = 0
At x=0x = 0, the yy-coordinate is 2sin⁡(0)−cos⁡(0)=−12\sin(0) - \cos(0) = -1. Evaluating the derivative dydx=2cos⁡x+sin⁡x\frac{dy}{dx} = 2\cos x + \sin x at x=0x = 0 yields a tangent slope of 22. Since the normal is perpendicular to the tangent, its gradient is −12-\frac{1}{2}. Substituting the point (0,−1)(0, -1) and slope −12-\frac{1}{2} into the line formula yields x+2y+2=0x + 2y + 2 = 0.

Adım Adım Çözüm

1
Find the yy-coordinate of the point of contact
At x=0x = 0, y=2sin⁡(0)−cos⁡(0)=0−1=−1y = 2\sin(0) - \cos(0) = 0 - 1 = -1. The point is (0,−1)(0, -1).
The line equation requires a point (x1,y1)(x_1, y_1) on the curve.
2
Differentiate the curve to find dydx\frac{dy}{dx}
dydx=2cos⁡x−(−sin⁡x)=2cos⁡x+sin⁡x\frac{dy}{dx} = 2\cos x - (-\sin x) = 2\cos x + \sin x.
The derivative gives the gradient function of the curve.
3
Calculate the gradient of the tangent and normal at x=0x = 0
Tangent gradient mt=2cos⁡(0)+sin⁡(0)=2(1)+0=2m_t = 2\cos(0) + \sin(0) = 2(1) + 0 = 2. Normal gradient mn=−1mt=−12m_n = -\frac{1}{m_t} = -\frac{1}{2}.
The normal line is perpendicular to the tangent line.
4
Form the equation of the normal line
y−(−1)=−12(x−0)  ⟹  y+1=−12x  ⟹  2y+2=−x  ⟹  x+2y+2=0y - (-1) = -\frac{1}{2}(x - 0) \implies y + 1 = -\frac{1}{2}x \implies 2y + 2 = -x \implies x + 2y + 2 = 0.
Apply the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1).

Anahtar Kavram

Equation of Normal to a Curve
Tahmini Süre:1m 30s
Soru 24Soru →

Calculate the gradient of the normal line to the curve y=6xy = \frac{6}{x} at the point where x=3x = 3.

Cevabı ve açıklamayı göster

Cevap: 1.5

Cevap

The gradient of the normal line is 1.5.
Differentiating y=6x−1y = 6x^{-1} yields dydx=−6x2\frac{dy}{dx} = -\frac{6}{x^2}. Evaluating this derivative at x=3x = 3 gives the tangent gradient mt=−69=−23m_t = -\frac{6}{9} = -\frac{2}{3}. Because the normal line is perpendicular to the tangent line, its gradient is the negative reciprocal mn=−1mt=32=1.5m_n = -\frac{1}{m_t} = \frac{3}{2} = 1.5.

Adım Adım Çözüm

1
Differentiate the function y=6x−1y = 6x^{-1} with respect to xx
dydx=−6x−2=−6x2\frac{dy}{dx} = -6x^{-2} = -\frac{6}{x^2}
The first derivative represents the formula for the tangent gradient to the curve at any given point.
2
Evaluate the derivative at x=3x = 3 to find the tangent slope (mtm_t)
m_t = -\frac{6}{3^2} = -\frac{6}{9} = -\frac{2}{3}
Substituting the given x-coordinate into the derivative gives the exact slope of the tangent at that point.
3
Calculate the normal slope (mnm_n) as the negative reciprocal of mtm_t
m_n = -\frac{1}{m_t} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} = 1.5
The normal line is perpendicular to the tangent line, meaning mt⋅mn=−1m_t \cdot m_n = -1.

Anahtar Kavram

The gradient of the normal to a curve y=f(x)y = f(x) at x=ax = a is the negative reciprocal of the derivative evaluated at that point: mn=−1f′(a)m_n = -\frac{1}{f'(a)}.
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