MHB Calculus Tangent Line: Find Parameter k | Camille's Q&A

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To find the parameter k for the curve y=x^3+k that is tangent to the line 3x-4y=0 in the first quadrant, the slope of the tangent line is determined to be 3/4. By equating this slope to the derivative of the curve, it is found that x=1/2 is the point of tangency. Substituting this x-value into both the curve and the tangent line reveals that k must equal 1/4 for the two to intersect at the same point. A plot confirms the tangent line and the curve intersect correctly at this value of k. Thus, the solution concludes that k=1/4.
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Here is the question:

Calculus tangent line?

If the line 3x-4y=0 is tangent in the first quadrant to the curve y=x^3+k then k is
(a)1/2
(b)1/4
(c)0
(d)-1/8
(e)-1/2
Could you please show work because I am so lost

I have posted a link there to this topic so the OP may see my work.
 
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Hello Camille,

Let's begin by equating the derivative of the curve to the slope of the tangent line. The tangent line may be written as $$y=\frac{3}{4}x$$ hence:

$$\frac{3}{4}=3x^2$$

$$x^2=\frac{1}{4}$$

Since we are interesting in the first quadrant solution, we take the positive root:

$$x=\frac{1}{2}$$

Now, the tangent line and the curve have then tangent point in common, and so using the tangle line, we know this point is:

$$\left(\frac{1}{2},\frac{3}{4}\cdot\frac{1}{2} \right)=\left(\frac{1}{2},\frac{3}{8} \right)$$

Hence, we must have:

$$y\left(\frac{1}{2} \right)=\left(\frac{1}{2} \right)^3+k=\frac{1}{8}+k=\frac{3}{8}$$

And so we must have:

$$k=\frac{1}{4}$$

Here is a plot of the curve $$y=x^3+\frac{1}{4}$$ and the tangent line $$y=\frac{3}{4}x$$ for $0\le x\le1$:

View attachment 1322
 

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