AN EQUATION OF THE LINE NORMAL TO THE GRAPH OF
Y= SQRT[3X^2 + 2X] AT (2, 4) IS??
CALCULUS - Damon, Friday, January 2, 2009 at 5:23pm
y = (3 x^2 + 2 x)^.5
dy/dx = .5 [(3x^2+2x)^-.5 ] (6 x +2)
at (2,4), x = 2
dy/dx = .5 [ (3*4+4)^-.5 ] (14)
= .5 [ 1/4 ] (14)
= 7/4
the slope of the normal to that is -4/7
so
y = -(4/7) x + b
now at (2,4)
4 = -(4/7)*2 + b
4 = -8/7 + b
28/7 = -8/7 + b
b = 36/7
so
7 y = -4 x + 36
y' = (3x+1)/(3x^2+2x)
So the slope of the tangent at (2,4) is
7/4
so the slope of the normal is -4/7
let the equation of the normal be
4x + 7y = C, then at (2,4)
8 + 28 = C = 36
the equation of the normal is 4x + 7y = 36
y' should have bee
y' = (3x+1)/(3x^2+2x)^(1/2)
my calculations reflected the correct version
Well, my answer goes through (2,4) now :)
Well, well, well! Talking about lines and graphs, huh? Alright, let's have some fun solving this problem!
First, let's find the derivative of the given function. Taking the derivative of y with respect to x, we get:
y' = (1/2) * (3x^2 + 2x)^(-1/2) * (6x + 2)
Now, in order to get the equation of the line that is normal to the graph, we need the slope of the line. The slope of a line normal to another line is the negative reciprocal of the slope of that line.
So, let's find the slope of the tangent line at x = 2. We substitute x = 2 into the derivative:
y' = (1/2) * (3(2)^2 + 2(2))^(-1/2) * (6(2) + 2)
= (1/2) * (12 + 4)^(-1/2) * (12 + 2)
= (1/2) * (16)^(-1/2) * 14
= 7/4
Now, since we want the slope of the line normal to the graph, we take the negative reciprocal of 7/4, which gives us -4/7.
We know the slope (-4/7) and a point (2, 4) on the line, which means we can use the point-slope form to find the equation of the line.
Using the point-slope form, we have:
y - y1 = m(x - x1)
y - 4 = (-4/7)(x - 2)
y - 4 = (-4/7)x + (8/7)
y = (-4/7)x + (8/7) + 4
y = (-4/7)x + (8/7) + (28/7)
y = (-4/7)x + 36/7
And there you have it! The equation of the line normal to the graph of y = sqrt[3x^2 + 2x] at (2, 4) is y = (-4/7)x + 36/7. Keep it cool!
To find the equation of the line that is normal to the graph of a given function, we need to follow a few steps:
Step 1: Find the derivative of the given function.
Step 2: Evaluate the derivative at the given point to find the slope of the tangent line.
Step 3: Find the negative reciprocal of the slope from Step 2 to get the slope of the normal line.
Step 4: Use the point-slope form of a line to determine the equation of the normal line.
Let's follow these steps to find the equation of the line normal to the graph of \(y = \sqrt{3x^2 + 2x}\) at the point (2, 4).
Step 1: Find the derivative of \(y\) with respect to \(x\).
To find the derivative, we can use the power rule and the chain rule. The derivative of \(\sqrt{3x^2 + 2x}\) can be calculated as follows:
\[
\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt{3x^2 + 2x}}} \cdot (6x + 2)
\]
Step 2: Evaluate the derivative at the given point (2, 4) to find the slope of the tangent line.
Substitute \(x = 2\) into the derivative we found in Step 1:
\[
\frac{{dy}}{{dx}} \bigg|_{x=2} = \frac{1}{{2\sqrt{3\cdot 2^2 + 2\cdot 2}}} \cdot (6\cdot 2 + 2) = \frac{10}{\sqrt{28}} = \frac{5\sqrt{7}}{7}
\]
Therefore, the slope of the tangent line to the graph of \(y = \sqrt{3x^2 + 2x}\) at the point (2, 4) is \(\frac{5\sqrt{7}}{7}\).
Step 3: Find the negative reciprocal of the slope from Step 2 to get the slope of the normal line.
The slope of the normal line will be the negative reciprocal of the slope of the tangent line. So, the slope of the normal line is given by:
\[
m_{\text{{normal}}} = -\frac{1}{{\frac{5\sqrt{7}}{7}}} = -\frac{7}{{5\sqrt{7}}}
\]
Step 4: Use the point-slope form of a line to determine the equation of the normal line.
We can use the point-slope form of a line, which is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the given point, and \(m\) is the slope of the normal line.
Substituting \((x_1, y_1) = (2, 4)\) and \(m = -\frac{7}{{5\sqrt{7}}}\) into the point-slope form, we get:
\[
y - 4 = -\frac{7}{{5\sqrt{7}}}(x - 2)
\]
We can simplify this equation by multiplying both sides by \(5\sqrt{7}\):
\[
5\sqrt{7} \cdot (y - 4) = -7(x - 2)
\]
This gives us the final equation of the line normal to the graph of \(y = \sqrt{3x^2 + 2x}\) at the point (2, 4):
\[
5\sqrt{7}y - 20\sqrt{7} = -7x + 14
\]
y = (3 x^2 + 2 x)^.5
dy/dx = .5 [(3x^2+2x)^-.5 ] (6 x +2)
at (2,4), x = 2
dy/dx = .5 [ (3*4+4)^-.5 ] (14)
= .5 [ 1/4 ] (14)
= 7/4
the slope of the normal to that is +4/7
so
y = (4/7) x + b
now at (2,4)
4 = (4/7)*2 + b
4 = 8/7 + b
28/7 = 8/7 + b
b = 20/7
so
7 y = 4 x + 20