The line x=c where c>0 intersects the cubic y=2x^(3)+3x^(2)-9 at point P and the parabola y=4x^(2)+4x+5 at point Q.
a. If a line tangent to the cubic at point P is parallel to the line tangent to the parabola at point Q, find the value of c where c>0.
See:
http://www.jiskha.com/display.cgi?id=1255564276
Oh, c'mon, let's solve this problem and make some "parallel parking" jokes along the way! Here we go:
First, let's find the derivative of the cubic function:
y = 2x^3 + 3x^2 - 9
dy/dx = 6x^2 + 6x
To find the slope of the line tangent to the cubic at point P, we need to substitute x = c into the derivative:
m1 = 6c^2 + 6c
Now, let's find the derivative of the parabola:
y = 4x^2 + 4x + 5
dy/dx = 8x + 4
To find the slope of the line tangent to the parabola at point Q, we need to substitute x = c into the derivative:
m2 = 8c + 4
Since we want the line tangent to the cubic to be parallel to the line tangent to the parabola, their slopes should be equal. Therefore, we have:
6c^2 + 6c = 8c + 4
Simplifying this equation gives us:
6c^2 - 2c - 4 = 0
Now, let's use some funny quadratic formula to solve this equation:
c = (-b ± √(b^2 - 4ac)) / (2a)
Doing some math, we find:
c = (-(-2) ± √((-2)^2 - 4(6)(-4))) / (2(6))
c = (2 ± √(4 + 96)) / 12
c = (2 ± √100) / 12
c = (2 ± 10) / 12
So, we have two possible values for c:
c1 = (2 + 10) / 12 = 12 / 12 = 1
c2 = (2 - 10) / 12 = -8 / 12 = -2/3
Since we're given c > 0, the only valid solution is:
c = 1
So, the value of c where the line tangent to the cubic at point P is parallel to the line tangent to the parabola at point Q is c = 1. And hey, don't forget to parallel park your answer nicely!
To find the value of c where the line tangent to the cubic at point P is parallel to the line tangent to the parabola at point Q, we need to find the slopes of both tangents.
1. The cubic function y = 2x^3 + 3x^2 - 9.
The slope of the tangent to the cubic at point P(x, y) is given by the derivative dy/dx.
Taking the derivative of the cubic function:
dy/dx = 6x^2 + 6x
2. The parabolic function y = 4x^2 + 4x + 5.
The slope of the tangent to the parabola at point Q(x, y) is again given by the derivative dy/dx.
Taking the derivative of the parabola function:
dy/dx = 8x + 4
Now, we know that the slopes of the two tangents are equal since they are parallel. So, we can equate the derivatives and solve for x:
6x^2 + 6x = 8x + 4
Simplifying the equation:
6x^2 - 2x - 4 = 0
We can now solve this quadratic equation to find the values of x. Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values:
a = 6, b = -2, c = -4
x = (-(-2) ± √((-2)^2 - 4(6)(-4))) / (2(6))
x = (2 ± √(4 + 96)) / 12
x = (2 ± √100) / 12
x = (2 ± 10) / 12
Simplifying further, we have two possible values for x:
x = (2 + 10) / 12 = 12 / 12 = 1 (1)
x = (2 - 10) / 12 = -8 / 12 = -2/3 (-2/3)
Now, since we are looking for a positive value of c, we can discard the second value of x (-2/3) since it is negative.
Therefore, the value of c where the line tangent to the cubic at point P is parallel to the line tangent to the parabola at point Q is c = 1.
To find the value of c where the line tangent to the cubic at point P is parallel to the line tangent to the parabola at point Q, we can start by finding the slopes of the tangents at points P and Q.
Let's begin by finding the derivative of the cubic function y=2x^(3)+3x^(2)-9. The derivative will give us the slope of the tangent line at any point on the cubic.
dy/dx = d/dx (2x^(3)+3x^(2)-9)
Using the power rule for differentiation, we can find the derivative:
dy/dx = 6x^(2) + 6x
Now let's find the derivative of the parabola function y=4x^(2)+4x+5:
dy/dx = d/dx (4x^(2)+4x+5)
= 8x + 4
To find the slope at point P on the cubic, we substitute the value c into the derivative of the cubic:
dy/dx = 6c^(2) + 6c
Similarly, to find the slope at point Q on the parabola, we substitute the value c into the derivative of the parabola:
dy/dx = 8c + 4
Since the line tangent at P is parallel to the line tangent at Q, their slopes must be equal:
6c^(2) + 6c = 8c + 4
Simplifying the equation, we get:
6c^(2) - 2c - 4 = 0
Now we can solve this quadratic equation for c. We can use the quadratic formula:
c = (-b ± √(b^(2)-4ac))/(2a)
In this case, a = 6, b = -2, and c = -4. Substituting these values into the quadratic formula, we have:
c = (-(-2) ± √((-2)^(2)-4(6)(-4)))/(2(6))
= (2 ± √(4+96))/(12)
= (2 ± √100)/(12)
= (2 ± 10)/(12)
Simplifying further, we get:
c = (2 + 10)/12 = 12/12 = 1
or
c = (2 - 10)/12 = -8/12 = -2/3
Since c must be greater than zero (c > 0) according to the given condition, the value of c that satisfies the condition is c = 1.