Finding Value of k for Tangent to y=\frac{3x}{\sqrt{1+x}} at x=3

In summary, the formula for finding the value of k for tangent to y=\frac{3x}{\sqrt{1+x}} at x=3 is k = \frac{dy}{dx}|_{x=3}. To calculate the derivative of y=\frac{3x}{\sqrt{1+x}}, we can use the quotient rule: \frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}. The value of \frac{dy}{dx} at x=3 for y=\frac{3x}{\sqrt{1+x}} can be found by plugging in x=3
  • #1
icystrike
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Homework Statement


The equation of a curve is y=[tex]\frac{3x}{\sqrt{1+x}}[/tex].
Given that the equation of the tangent to the curve at the point x=3 us 15x-16y=k , find the value of k. SOLVED

Homework Equations


The Attempt at a Solution

 
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  • #2
Do you know the y-coordinate on the curve when x = 3?
Do you know how to find the slope of a curve (i.e., the slope of the tangent line to the curve)?
Do you know how to find the slope of the line 15x - 16y = k?
 
  • #3


To find the value of k, we can use the formula for the slope of a tangent line, which is equal to the derivative of the curve at the given point.

First, we need to find the derivative of the curve y=\frac{3x}{\sqrt{1+x}}. Using the power rule and the quotient rule, we get:

y'=\frac{3\sqrt{1+x}-3x\frac{1}{2}(1+x)^{-\frac{1}{2}}}{1+x}

Next, we can substitute x=3 into the derivative to find the slope of the tangent line at x=3:

y'=\frac{3\sqrt{4}-3(3)\frac{1}{2}(4)^{-\frac{1}{2}}}{1+4}

y'=\frac{3\sqrt{4}-3(3)\frac{1}{2}(2)}{5}

y'=\frac{3(2)-\frac{9}{2}}{5}

y'=\frac{3}{2}

Now, we can use the point-slope form of a line to find the equation of the tangent line at x=3:

y-y_1=m(x-x_1)

y-(\frac{9}{\sqrt{13}})=\frac{3}{2}(x-3)

y-\frac{9}{\sqrt{13}}=\frac{3}{2}x-\frac{9}{2}

y=\frac{3}{2}x+\frac{9}{2\sqrt{13}}

Finally, we can compare the equation of the tangent line to the given equation 15x-16y=k and see that k=\frac{9}{2\sqrt{13}}. Therefore, the value of k is approximately 0.65.
 

1. What is the formula for finding the value of k for tangent to y=\frac{3x}{\sqrt{1+x}} at x=3?

The formula for finding the value of k for tangent to y=\frac{3x}{\sqrt{1+x}} at x=3 is k = \frac{dy}{dx}|_{x=3}.

2. How do you calculate the derivative of y=\frac{3x}{\sqrt{1+x}}?

To calculate the derivative of y=\frac{3x}{\sqrt{1+x}}, we can use the quotient rule: \frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}. In this case, u=3x and v=\sqrt{1+x}. Plugging in these values, we get \frac{d}{dx}(\frac{3x}{\sqrt{1+x}}) = \frac{\sqrt{1+x} - 3x\frac{1}{2\sqrt{1+x}}}{1+x} = \frac{1-3x}{2(1+x)^{\frac{3}{2}}}.

3. What is the value of \frac{dy}{dx} at x=3 for y=\frac{3x}{\sqrt{1+x}}?

The value of \frac{dy}{dx} at x=3 for y=\frac{3x}{\sqrt{1+x}} can be found by plugging in x=3 into the derivative formula: \frac{dy}{dx}|_{x=3} = \frac{1-3(3)}{2(1+3)^{\frac{3}{2}}} = \frac{-8}{16\sqrt{4}} = -\frac{1}{2\sqrt{4}} = -\frac{1}{4}.

4. How do you find the value of k for tangent to y=\frac{3x}{\sqrt{1+x}} at x=3 using the derivative?

To find the value of k for tangent to y=\frac{3x}{\sqrt{1+x}} at x=3 using the derivative, we simply plug in x=3 into the derivative formula: k = \frac{dy}{dx}|_{x=3} = -\frac{1}{4}.

5. Can the value of k for tangent to y=\frac{3x}{\sqrt{1+x}} at x=3 be negative?

Yes, the value of k for tangent to y=\frac{3x}{\sqrt{1+x}} at x=3 can be negative. This means that the slope of the tangent line at x=3 is negative, indicating that the curve is decreasing at that point.

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