第6章衍生物的应用-练习6.3 |套装1
问题14:在指定点找到给定曲线的切线和法线的方程式:
(i)在(0,5)处y = x 4 – 6x 3 + 13x 2 – 10x + 5
(ii)在(1,3)处y = x 4 – 6x 3 + 13x 2 – 10x + 5
(iii)在(1,1)处y = x 3
(iv)y = x 2 at(0,0)
(v)x = cos t,y =在t =π/ 4时的sin t
解决方案:
(i) Given curve
y = x4 – 6x3 + 13x2 – 10x + 5
Given point, (0, 5)
dy/dx = 4x3 – 18x2 + 26x – 10
dy/dx = 4(0)3 – 18(0)2 + 26(0) – 10
dy/dx = -10,
-dx/dy = 1/10
Now, with the help of points slope form
y – y1 = m(x – x1)
y – 5 = -10(x – 0)
y + 10x = 5 is the required equation of the tangent
For equation of normal,
y – y1 =
y – 5 =
10y – x – 50 is the equation of normal.
(ii) Given curve: y = x4 – 6x3 + 13x2 – 10x + 5
Given point is (1, 3)
dy/dx = 4x3 – 18x2 + 26x – 10
dy/dx = 4(1)3 – 18(1)2 + 26(1) – 10
dy/dx = 4 – 18 + 26 – 10 = 2
dy/dx = 2
-dx/dy = -1/2
Using point slope form, equation of tangent is
y – y1 = dy/dx(x – x1)
y – 3 = 2(x – 1)
y – 2x = 1 is the equation of tangent.
Using point slope form, equation of normal is
y – y1 = -dx/dy(x – 1)
y – 3 = -1/2(x – 1)
2y – 6 = -x + 1
2y + x = 7 is the equation of normal.
(iii) Given curve : y = x3
Given point is (1, 1)
dy/dx = 3x2
dy/dx = 3(1)2 = 3
dy/dx = 3 & -dx/dy = -1/3
Using point slope form, equation of tangent is y – y1 = dy/dx(x – x1)
y – y1 = dy/dx(x – x1)
y – 1 = 3(x – 1)
y – 3x + 2 = 0 is the equation of tangent
Using point slope form, equation of normal is
y – y1 =
y – 1 =
3y – 3 = -x + 1
3y + x = 4 is the equation of normal.
(iv) Given curve: y = x2
Given point (0, 0)
dy/dx = 2x
dy/dx = 0
dy/dx = 0 & -dx/dy = not defined is
y – y1 = dy/dx(x – x1)
y – 0 = 0(x – 0)
y = 0 is the equation of tangent.
Using point slope form, equation of normal is
y – y1 = -(1)
-dx\dy is undefined, so we can write eq(1) as
Now putting dy/dx = 0 we get
0(y – 0) = x-0
x = 0 is the equation of normal.
(v) Equation of curve: x = cos t and y = sin t
Point t = π/4
-(1)
On putting these values in eq(1), we get
dy/dx = -1 & -dx/dy = 1
Now for t = π/4,
y1 = sin t = sin(π/4) = 1/√2
x1 = cos t = cos(π/4) = 1/√2
The point is (1/√2, 1/√2)
y – y1 =
y – (1/√2) = -1(x – 1/√2)
y – 1/√2 = -x + 1/√2
x + y = √2 is the equation of normal is
y – 1/√2 = 1(x – 1/√2)
x = y is the equation of normal.
问题15.求曲线y = x 2 – 2x + 7的切线的方程为
(i)平行于第2x – y + 9 = 0行
(ii)垂直于线5y – 15x = 13
解决方案:
Given curve: y = x2 – 2x + 7
On differentiating w.r.t. x, we get
dy/dx = 2x – 2 -(1)
(i) Tangent is parallel to 2x – y + 9 = 0 that means,
Slope of tangent = slope of 2x – y + 9 = 0
y = 2x + 9
Slope = 2 -(Comparing with y = mx + e)
dy/dx = slope = 2
2x – 2 = 2
x1 = 2
Corresponding to x1 = 2,
y1 = x12 – 2x1 + 7
y1 = (2)2 – 2(2) + 7
y1 = 7
The point of contact is (2, 7).
Using point slope form, equation of tangent is
y – y1 =
y – 7 = 2(x – 2)
y – 2x = 3 is the equation of tangent.
(ii) Tangent is perpendicular to the line 5y – 15x = 13
That means (slope of tangent) x (slope of line) = -1
For, slope of line 5y – 15x = 13
5y = 15x + 13
y = 3x + 13/15
Slope = 3
(Slope of tangent) x 3 = -1
Slope of tangent =-1/3
Now, y = x2 – 2x + 7
dy/dx = 2x – 2
On comparing dy/dx with slope, we get
2x – 2 = -1/3
6x – 6 = -1
6x = 5
x1 = 5/6
For x1 = 5/6,
y1 = x12 – 2x1 + 7
y1 = (5/6)12 – 2(5/6)1 + 7
y1 = 217/36
Now using point slope form, equation of tangent is
y – y1 = m(x – x1)
36y – 217 = -12x + 10
36y + 12x = 227 is the required equation of tangent.
问题16:证明在x = 2和x = -2的点处,曲线y的切线y = 7x 3 + 11是平行的。
解决方案:
Given curve: y = 7x3 + 11
On differentiating w.r.t. x, we get
dy/dx = 21x2
dy/dx = 21(2)2 = 84
The slopes at x – 2 & -2 are the same,
Hence the tangent will be parallel to each other.
问题17.在曲线y = x 3上找到切线的斜率等于该点的y坐标的点。
解决方案:
Given curve: y = x3
On differentiating w.r.t. x, we get
dy/dx = 3x2 -(1)
Now let us assume that the point is (x1, y1)
dy/dx = 3x12
Also, slope of tangent at (x1, y1) is equal to y1.
So, 3x12 = y1 -(2)
Also, (x1, y1) lies on y = x3 x3,so
y1 = x13 -(3)
From eq(2) & (3)
3x12 = x13
3x12 = x13 = 0
x12(3 – x1) = 0
For x1 = 0, y1 = x13 = (0)3 = 0
One such point is (0, 0)
For x1 = 3, y1. = (3)3 = 27
Second point is (3, 27)
问题18.对于曲线y = 4x 3 – 2x 5 ,找到切线穿过原点的所有点。
解决方案:
Given curve: y = 4x3 – 3x5
Clearly at x = 0, y = 0, i.e the curve passes through origin.
Now the tangent also passes through origin.
Equation of a line passing through origin is y = mx.
Now tangent is touching the curve, so
y = mx will satisfy in curve.
mx = 4x3 – 2x5 -(1)
Now dy/dx = 12x2 – 10x4
Also m is the slope of tangent, so
m = 12x2 – 10x4 -(2)
From eq(1) & (2),
(12x2 – 10x4)x = 4x3 – 3x5
x3(12 – 10x2) = x3(4 – 2x2) -(3)
For the first point, x = 0
For x1 = 0, y1 = 4x13 – 2x15 = 0
So, (0, 0) is one such point
Now for other roots of 3
12 – 10x2 = 4 – 2x2
8 = 8x2
x2 = 1
x = ±1
For x2 = 1, y2 = 4x23 – 2x25 = 4(1)3 – 2(1)5 = 2
For x3 = -1, y3 = 4x33 – 2x35 = 4(-1)3 – 2(-1)5 = -2
The other points are(1, 2) & (-1, -2)
问题19.在曲线x 2 + y 2 – 2x – 3 = 0上找到切线平行于x轴的点。
解决方案:
Given curve: x2 + y2 – 2x – 3 = 0
On differentiating w.r.t. x, we get
3x + 2y(dy/dx) – 2 – 0 = 0
Given that the tangent are parallel to x-axis,
So, dy/dx = slope = 0
1 – x/y = 0
For x1 = 1,
x12 + y12 – 2x1 – 3 = 0
(1)2 + y12 – 2(1) – 3 = 0
y12 = 4
y12 = ≠2
The points are (1, 2) & (1, -2)
问题20:找到曲线ay 2 = x 3的点(am 2 ,am 3 )的法线方程。
解决方案:
Given curve: ay2 = x3
On differentiating w.r.t. x, we get
2ay.dy/dx = 3x2
&
So, by point slope form, equation of normal is,
3my – 3am4 = -2x + 2am2
Hence, 3my + 2x = 2am2 + 3am4 is the required equation of normal.
问题21.找到与直线x + 14y + 4 = 0平行的曲线y = x 3 + 2x + 6的法线方程。
解决方案:
Given curve: y = x3 + 2x + 6
On differentiating w.r.t. x, we get
-(Slope of normal)
Now, the normal are parallel to x + 14y + 4 = 0
13 = 3x2 + 2
3x2 = 12 ⇒ x2 = 4
x = ±2
x1 = 2 & x2 = -2
For x1 = 2; y1 = = (2)3 + 2(2) + 6 = 18
For x2 = -2; y2 = = (-2)3 + 2(-2) + 6 = -6
Normal through (2,18) is
y – 18 =
14y – 252 = -x + 2
14y + x = 254 is one such equation.
Normal through (-2, -6) is
y + 6 =
14y + 84 = -x – 2
14y + x + 86 = 0 is the other equation of normal.
问题22查找切线和法线的方程来抛物线Y 2 = 4AX在点(在图2,2原子)。
解决方案:
Given parabola: y2 = 4ax
On differentiating w.r.t. x, we get
Now, by point slope form, equation of tangent is,
y – y1 =
ty = x + at2 is the equation of tangent to the parabola y2 = 4ax at (at2, 2at)
Now
Now, by point slope form, equation of normal is,
y – y1 =
y – 2at = -t(x – at2)
y + xt = 2at + at3 is the equation of normal to the parabola y2 = 4ax at (at2, 2at)
问题23.证明如果8k 2 = 1,曲线x = y 2和xy = k切成直角。
解决方案:
Given curves: x = y2 & xy = k
Two curves intersect at right angles f the tangents through their point
intersection is perpendicular to each other.
Now if tangents are perpendicular their product of their slopes will be equal to -1.
Curve 1: x = y2
1 = 2y.dy/dx
dy/dx = 1/2 y = m1 -(1)
Curve 2: xy = k
y = k/x
Let’s find their point of intersection
x = y2 & xy = k
k/y = y2
y = k1/3
x = y2
x = k2/3
The point is (k2/3, k1/3)
m1 = 1/2k1/3
For curves to be intersecting each other at right angles,
m1m2 = -1
8k2 = 1
Hence proved
问题24.找到双曲线的切线和法线的方程在点(x o ,y o )。
解决方案:
Given curve:
On differentiating both sides with respect to x,
Now,
Equation of tangent by point slope form is,
-((x0,y0) lie on )
is the equation of tangent.
Now,
Equation of normal by point slope form is,
x0b2y – x0b2y0 = -y0a2x + y0a2x0
x0b2y + y0a2x = x0y0(a2 + b2) is the equation of normal
问题25.求曲线y =的切线的方程。 与4x – 2y + 5 = 0线平行。
解决方案:
Given curve:
On differentiating w.r.t. x, we get
Now, it is given that the tangent is parallel to the line 4x – 2y + 5 = 0,
so their slopes must be equal.
Slope of 2y = 4x + 5 is 2.
So,
9/16 = 3x – 2
x1 = 41/48
Now,
The point is (41/48, 3/4)
Now by point slope form, the equation of tangent will be
y – y1 = m(x – x1)
24y – 48x + 23 = 0 is the required equation of tangent.
问题26.曲线y的法线y = 2x 2 + 3 sin x在x = 0处的斜率是
(A)3(B)1/3(C)-3(D)-1/3
解决方案:
Given curve: y = 2x2 + 3 sin x
On differentiating w.r.t. x, we get
dy/dx = 4x + 3cos x
-(Slope of normal)
Hence, the correct option is D.
问题27.线y = x +1与曲线y 2 = 4x在该点的切线
(A)(1,2)(B)(2,1)(C)(1,-2)(D)(-1,2)
解决方案:
Given curve: y2 = 4x
On differentiating w.r.t. x, we get
y = x + 1 is tangent, slope is 1, so, 2/y = 2
y1 = 2,
y12 = 4x1
x1 =
x1 = 1
So, (1, 2) is the point.
Hence, the correct option is A.