If the line 4x-9y=0 is tangent in the first quadrant to the graph of y =1/3x^3+c what is c?

Question

Charles Arocha · Accepted Answer

#c=16/81#

We'll have to do some thinking in order to solve this problem.

First of all, from a calculus point of view, what does "tangent" mean? Well, it's a line that touches a curve at one point. And what do we know about lines? They have constant slopes! But the question is asking us about #y=1/3x^3+c#, which is not a line - it's a curve. To find the slope of a curve, we need to take it's derivative, so we'll start there: #y=1/3x^3+c# #dy/dx=3*1/3x^(3-1)=x^2-># using the power rule

Ok, now where does that leave us? Well, we have an expression for the slope of the tangent line at a point (remember, that's what the derivative is). Wait a minute! Didn't the question tell us that the line #4x-9y = 0# is tangent to #y=1/3x^3+c#? In other words, the slope of the tangent line of #y=1/3x^3+c# is equal to the slope of #4x-9y=0# at the point where they are tangent (because #4x-9y=0# is the slope). If we add #9y# to both sides and divide by #9#, we see that #y=4/9x#. Let's set #4/9#, which is the slope, equal to #dy/dx# and see what happens: #4/9=x^2->x=2/3#

Note that we take the positive square root because the point in question is in the first quadrant, as specified by the question.

This means that #4x-9y=0# is tangent to #y=1/3x^3+c# at the point #x=2/3#. To find the #y#-value of this point, we plug in to #y=4/9x#: #y=4/9x=4/9*(2/3)=8/27#

So now we have a point, #(2/3, 8/27)#, and an unknown in #y=1/3x^3+c#. We can use our #x# and #y# values here to solve for #c#: #y=1/3x^3+c# #8/27=1/3(2/3)^3+c# #8/27=8/81+c# #c=8/27-8/81=16/81#

Samantha Adkison · Answer

undefined To find the value of c, we need to determine the point of tangency between the line 4x-9y=0 and the graph of y = 1/3x^3+c.

First, we need to find the derivative of the equation y = 1/3x^3+c, which is y' = x^2.

Next, we substitute the equation of the line into the derivative to find the x-coordinate of the point of tangency: 4x - 9y = 0 → 4x - 9(1/3x^3+c) = 0 → 4x - 3x^3 - 9c = 0.

Now, we solve this equation for x.

By factoring, we get: x(4 - 3x^2) = 9c.

Since the line is tangent in the first quadrant, the x-coordinate of the point of tangency must be positive.

Setting x = 0, we find that x = 0 is not a solution.

Setting 4 - 3x^2 = 0, we find that x = √(4/3) is a solution.

Therefore, the x-coordinate of the point of tangency is x = √(4/3).

To find the corresponding y-coordinate, we substitute this value of x into the equation y = 1/3x^3+c:

y = 1/3(√(4/3))^3 + c = 2√3/9 + c.

Since the line is tangent to the graph, the y-coordinate of the point of tangency must be equal to 0.

Setting 2√3/9 + c = 0, we find that c = -2√3/9.

Therefore, c = -2√3/9.