Problem #5: To find the equations of the lines tangent to the function
y=(x-1)/(x+1)
and parallel to the line with the equation
x-2y=3 ,
first find the slope of the line with this equation. Then, the tangent lines to the given curve will have the same slope, if they are parallel to the line.
The equation of the line is given in standard form. On of the ways to find its slope is to rewrite it in the slope-intercept form:
x-2y = 3
-2y = -x+3
y=1/2x -3/2
The slope is the coefficient in front of x, so the slope is
m=1/2
To determine which tangent lines will have the same slope, take the derivative of the given function using the quotient rule:
y' = ((x-1)/(x+1))' = ((x+1)-(x-1))/(x+1)^2 = 2/(x+1)^2
The derivative will equal 1/2 at the point where
2/(x+1)^2 = 1/2
(x+1)^2 = 4 (This is true assuming x cannot equal -1.)
Taking square root yields
x + 1 = 2 and x + 1 = -2. So the tangent line will have the needed slope at the points with x-coordinates x = 1 and x = -3.
The y-coordinates at these points are
y(1) = (1-1)/(1 + 1) = 0 and
y(-3) = (-3-1)/(-3+1) = (-4)/(-2) = 2
So the points are (1,0) and (-3, 2). The equations of the tangent lines can be found using the point-slope form. For the one of the lines,
y-0 = 1/2(x-1)
y = 1/2x -1/2
For the second line,
y-2=1/2(x-(-3))
y= 1/2x+3/2+2 = 1/2x +7/2
The first line has the equation
y = 1/2x - 1/2 (smaller intercept) and the second line has the equation
y=1/2x + 7/2 .
Problem #6:
a) Question 1: f(10) = 10,000 means that at the price p = $10 per yard, the amount of fabric sold is 10,000 yards. This is according to the definition of f(10) - the quantity sold as a function of price. This is choice 3.
Question 2: f'(10) = -325. The derivative of a function means its rate of change. Positive derivative means positive rate (increase), and negative derivative means negative rate (decrease). So the derivative of the function representing quantity indicates how fast the quantity increases or decreases. At the price of $10 per yard, the quantity decreases at the rate of 325 yards per (dollar per yard). This is choice 5.
b) To find R'(10), use the product rule to calculate the derivative of R(p):
R'(p) = (pf(p))' = p'f(p)+pf'(p) = f(p) + pf'(p)
For p = 10, use the values given in part a) and plug them in:
R'(10) = 10,000 + 10*(-325) = 10,000 - 3,250 = 6,750.
This is the rate of change of the revenue, which means that at the price of $10 per yard, the revenue is increasing at the rate of $6,750 per (dollar per yard). This is choice 5.
Problem #7: Find the equation of the tangent line to the function
y = 6x^2 - x^3
at the point (1, 5).
The slope of the tangent line is the value of the derivative at the given point. The derivative of the given function is
y' = 12x - 3x^2
At point x = 1, y'(1) = 12*1 - 3*1^2 = 9
The slope of the tangent line is m = 9. The equation of line with the given slope passing through the point (1, 5) can be found using the point-slope form:
y-5 = 9(x-1)
In the slope-intercept form, this becomes
y=9x-4
To determine which of the four given graphs correspond to this situation, consider that only top and bottom graphs on the right have the tangent line go through the point (1,5). The functions on these graphs look similar, but notice that the top function has local maximum and minimum at the points x = -4 and x = 0, respectively. The given function has the local extrema at the points x = 0 and x = 4 (that's where the derivative
y=12x - 3x^2
becomes zero), as the function on the bottom right graph.
The given function and its tangent line at (1,5) are graphed at the bottom right graph.
Problem #1. Find the points on the curve y = 2x^3+3x^2-12x+5 where the tangent line is horizontal.
Solution:
Take the derivative of the function.
y'=6x^2+6x-12
A horizontal line has a slope of zero. So set the derivative equal to zero. And solve for x.
0=6x^2+6x-12
0=6(x^2+x-2)
0=x^2+x-2
0=(x+2)(x-1)
x+2=0, x=-2
x-1=0, x=1
And, solve for y values. To do so, plug-in the x values to original function.
y = 2x^3+3x^2-12x+5
x=-2, y=2(-2)^3+3(-2)^2-12(-2)+5 = 25
x=1, y=2(1)^3+3(1)^2-12(1)+5=-2
Therefore, the tangent line is horizontal at points (-2,25) and (1,-2).
Problem #2. Where does the normal line to the parabola y=x-x^2 at the point (1,0) intersect the parabola a second time?
Solution:
The slope of the line tangent to the curve at (1,0) should be determined. So, take the derivative of the function.
y=x-x^2
y'=1-2x
Then, plug-in x=1.
y'=1-2(1)
y'=-1
So, the slope of the tangent line is m = -1.
The normal line is perpendicular to the tangent line. Hence, the slope of the normal line is m=1.
Then, apply the point-slope form to get the equation of the normal line.
y-y_1=m(x-x_1)
Plug-in the slope m=1 and the given point (1,0).
y-0=1(x-1)
y=x-1
This is the equation of the normal line.
Then, determine the points where the parabola and the normal line intersect. To do so, set the two ys equal to each other.
y=x - 1
y=x-x^2
x-1=x-x^2
Solve for the x values.
x^2-1=0
(x-1)(x+1)=0
x-1=0, x=1
x+1=0, x=-1
And plug-in the x values to the equation of parabola or to the equation of normal line.
y=x-1
x=1, y=1-1=0
x=-1, y=-1-1=-2
The intersection points of the parabola and normal line are (1,0) and (-1,-2).
Therefore, the second intersection point is (-1,-2).
Problem #3. Find an equation of the tangent line to the given curve at the specified point.
y=(x^2-1)/(x^2+x+1), (1,0)
Solution:
Take the derivative of the function. Apply the quotient rule.
y'= ((x^2+x+1)(2x) - (x^2-1)(2x+1))/((x^2+x+1)^2)
y'=((2x^3 +2x^2+2x)-(2x^3+x^2-2x-1))/((x^2+x+1)^2)
y'=(x^2+4x+1)/(x^2+x+1)^2
Then, plug-in x=1 to get the slope of the tangent line.
y'=(1^2+4(1)+1)/((1^2+1+1)^2)
y'=6/9
y'=2/3
So the tangent line has a slope of m=2/3 and passes the point (1,0).
To determine the equation of the line, apply the point-slope form.
y-y_1=m(x-x_1)
y-0=2/3(x-1)
y=2/3x-2/3
Therefore, the equation of the tangent line is y=2/3x-2/3 .
Problem #4. If H(theta)= theta cos(theta) , find H'(theta) and H"(theta) .
Solution:
The first derivative of the function is:
H'(theta) = 1*cos(theta) + theta*(-sin(theta))
H'(theta) = cos(theta) - thetasin(theta)
The second derivative of the function is:
H"(theta) = -sin(theta) - (1*sin(theta) + theta*cos(theta))
H"(theta)=-2sin(theta)-thetacos(theta)
No comments:
Post a Comment