For what values of a and b is the line -4x+y=b tangent to the curve y=ax^3 when x=4?
Can anyone give me the hint?
dy/dx = 3 ax^2
when x = 4, dy/dx= 48 a = slope
but y = 4 x + b has slope of 4
so
48 a = 4
a = 4/48 = 1/12
and
y = (1/12) x^3
when x = 4
y = 16/3
-4(4) + 16/3 = b
-16 + 16/3 = b
b = -32/3
so
you can verify this at
http://www.wolframalpha.com/input/?i=plot+y%3D1%2F12+x^3,+y%3D4x-32%2F3,+0%3C%3Dx%3C%3D6
Sure, here's a hint for you: "When is a clown not funny? When it has no sense of tangent!"
Sure! To find the values of a and b for which the line -4x+y=b is tangent to the curve y=ax^3 when x=4, we need to use the concept of tangent lines and find the point of tangency.
Here's a step-by-step guide on how to approach this problem:
Step 1: Find the derivative of the curve y=ax^3.
- The derivative of y=ax^3 with respect to x gives us the slope of the tangent line to the curve at any given point.
Step 2: Find the slope of the tangent line at x=4.
- Plug in x=4 into the derivative obtained in Step 1 to find the slope of the tangent line at x=4.
Step 3: Set the slope of the tangent line equal to the slope of the given line.
- Since the line -4x+y=b is tangent to the curve at x=4, the slopes of both lines should be equal.
Step 4: Solve for a and b.
- Equate the slope of the tangent line (found in Step 2) to the slope of the line -4x+y=b and solve the resulting equation for a and b.
By following these steps, you will be able to find the values of a and b that make the line -4x+y=b tangent to the curve y=ax^3 when x=4.