For what values of a and b is the line –2x+y=b tangent to the curve y=ax^3 when x=–2?
a= ?
b= ?
Thanks guys!
First, you rewrite the equation as:
-2∙-2 + a∙-2^3
Next, you solve some of it to get:
4 + a∙8
Finally, if you can't figure out what a and b are, you can do what I did above, and try figuring it out. I did not know how to solve to get a and b, but at least you have the things above to help.
Oh, and by the way, it is actually Calculus instead of Caculus for the subject name of your problem. I just noticed it.
For a if it is 4, you will get the next equation of:
4 + 4∙8 = b
Finally, you solve, and you get:
36 as your answer for b.
So, a equals 4 and b equals 36.
Hint: you can always estimate values for a and then you solve, and whatever you get as your answer will be b.
y = ax^3
y' = 3ax^2
at x=-2, y' = 12a
The line y=2x+b has slope 2, so
12a=2
a = 1/6
at x = -2, y = -8a = -4/3 so the line must go through (-2,-4/3)
So, the tangent line has slope 2 and goes through (-2,-4/3)
y = 2x+b
-4/3 = -4 + b
b = 8/3
So, now we have
y = 2x - 8/3
is tangent to
y = 1/6 x^3
at (-2,-4/3)
To confirm this, visit
http://www.wolframalpha.com/input/?i=plot+y+%3D+1%2F6+x^3+and+y+%3D+2x%2B8%2F3+for+x+%3D+-4+..+4
typo near the end
y = 2x + 8/3
To determine the values of a and b for which the line -2x + y = b is tangent to the curve y = ax^3 when x = -2, we need to find the slope of the line and the slope of the curve at the given point.
1. Start by finding the slope of the line:
The equation -2x + y = b can be rearranged to y = 2x + b. Comparing this equation to the standard form y = mx + c, we can see that the slope of the line is m = 2.
2. Next, find the slope of the curve at x = -2:
We are given the equation y = ax^3. To find the slope of the curve at a particular point, we need to take the derivative of the equation with respect to x.
Differentiating y = ax^3, we get dy/dx = 3ax^2.
Substituting x = -2 into the derivative, we have:
dy/dx = 3a(-2)^2 = 3a(4) = 12a.
Therefore, the slope of the curve at x = -2 is 12a.
3. Since the line and the curve are tangent at x = -2, their slopes should be equal:
2 (slope of the line) = 12a (slope of the curve).
Setting these two equal, we get:
2 = 12a.
4. Solve for a:
Divide both sides of the equation by 12:
2/12 = a.
Simplifying, we have:
1/6 = a.
Therefore, the value of a is 1/6.
5. Substitute the value of a into the equation of the line to find b:
Using y = 2x + b, and substituting x = -2, we have:
-2(-2) + b = -4 + b = b.
Therefore, the value of b is -4.
In summary, the values of a and b for which the line -2x + y = b is tangent to the curve y = ax^3 when x = -2 are:
a = 1/6
b = -4.