Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
To find the equations of the tangent and the normal to the curve at the point where [tex]\( x = 2 \)[/tex], we will proceed step-by-step.
The curve is given by the equation:
[tex]\[ y = \frac{8}{x} - x + 3x^2 \][/tex]
First, let's find the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] to determine the slope of the tangent line at the given point. The derivative of [tex]\( y \)[/tex] is:
[tex]\[ y' = -\frac{8}{x^2} - 1 + 6x \][/tex]
We need to evaluate this derivative at [tex]\( x = 2 \)[/tex] to get the slope of the tangent line at that point:
[tex]\[ y'(2) = -\frac{8}{2^2} - 1 + 6 \cdot 2 \][/tex]
[tex]\[ y'(2) = -\frac{8}{4} - 1 + 12 \][/tex]
[tex]\[ y'(2) = -2 - 1 + 12 \][/tex]
[tex]\[ y'(2) = 9 \][/tex]
So, the slope of the tangent line at [tex]\( x = 2 \)[/tex] is 9.
Next, we need to find the [tex]\( y \)[/tex]-coordinate of the curve at [tex]\( x = 2 \)[/tex]. We substitute [tex]\( x = 2 \)[/tex] into the original equation:
[tex]\[ y(2) = \frac{8}{2} - 2 + 3 \cdot 2^2 \][/tex]
[tex]\[ y(2) = 4 - 2 + 12 \][/tex]
[tex]\[ y(2) = 14 \][/tex]
So, the point of tangency is [tex]\((2, 14)\)[/tex].
Now, the equation of the tangent line in point-slope form is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( m \)[/tex] is the slope of the tangent, and [tex]\((x_1, y_1)\)[/tex] is the point of tangency.
Substituting [tex]\( m = 9 \)[/tex], [tex]\( x_1 = 2 \)[/tex], and [tex]\( y_1 = 14 \)[/tex], we get:
[tex]\[ y - 14 = 9(x - 2) \][/tex]
[tex]\[ y - 14 = 9x - 18 \][/tex]
[tex]\[ y = 9x - 4 \][/tex]
So, the equation of the tangent line is:
[tex]\[ y = 9x - 4 \][/tex]
Next, we determine the equation of the normal line. The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is:
[tex]\[ m_{\text{normal}} = -\frac{1}{9} \][/tex]
Using the point-slope form again, the equation of the normal line at the point [tex]\((2, 14)\)[/tex] is:
[tex]\[ y - 14 = -\frac{1}{9}(x - 2) \][/tex]
[tex]\[ y - 14 = -\frac{1}{9}x + \frac{2}{9} \][/tex]
[tex]\[ y = -\frac{1}{9}x + \frac{2}{9} + 14 \][/tex]
[tex]\[ y = -\frac{1}{9}x + \frac{2}{9} + \frac{126}{9} \][/tex]
[tex]\[ y = -\frac{1}{9}x + \frac{128}{9} \][/tex]
So, the equation of the normal line is:
[tex]\[ y = -\frac{1}{9}x + \frac{128}{9} \][/tex]
In summary, the equations of the tangent and normal to the curve [tex]\( y = \frac{8}{x} - x + 3x^2 \)[/tex] at the point where [tex]\( x = 2 \)[/tex] are:
- Tangent line: [tex]\( y = 9x - 4 \)[/tex]
- Normal line: [tex]\( y = -\frac{1}{9}x + \frac{128}{9} \)[/tex]
The curve is given by the equation:
[tex]\[ y = \frac{8}{x} - x + 3x^2 \][/tex]
First, let's find the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] to determine the slope of the tangent line at the given point. The derivative of [tex]\( y \)[/tex] is:
[tex]\[ y' = -\frac{8}{x^2} - 1 + 6x \][/tex]
We need to evaluate this derivative at [tex]\( x = 2 \)[/tex] to get the slope of the tangent line at that point:
[tex]\[ y'(2) = -\frac{8}{2^2} - 1 + 6 \cdot 2 \][/tex]
[tex]\[ y'(2) = -\frac{8}{4} - 1 + 12 \][/tex]
[tex]\[ y'(2) = -2 - 1 + 12 \][/tex]
[tex]\[ y'(2) = 9 \][/tex]
So, the slope of the tangent line at [tex]\( x = 2 \)[/tex] is 9.
Next, we need to find the [tex]\( y \)[/tex]-coordinate of the curve at [tex]\( x = 2 \)[/tex]. We substitute [tex]\( x = 2 \)[/tex] into the original equation:
[tex]\[ y(2) = \frac{8}{2} - 2 + 3 \cdot 2^2 \][/tex]
[tex]\[ y(2) = 4 - 2 + 12 \][/tex]
[tex]\[ y(2) = 14 \][/tex]
So, the point of tangency is [tex]\((2, 14)\)[/tex].
Now, the equation of the tangent line in point-slope form is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( m \)[/tex] is the slope of the tangent, and [tex]\((x_1, y_1)\)[/tex] is the point of tangency.
Substituting [tex]\( m = 9 \)[/tex], [tex]\( x_1 = 2 \)[/tex], and [tex]\( y_1 = 14 \)[/tex], we get:
[tex]\[ y - 14 = 9(x - 2) \][/tex]
[tex]\[ y - 14 = 9x - 18 \][/tex]
[tex]\[ y = 9x - 4 \][/tex]
So, the equation of the tangent line is:
[tex]\[ y = 9x - 4 \][/tex]
Next, we determine the equation of the normal line. The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is:
[tex]\[ m_{\text{normal}} = -\frac{1}{9} \][/tex]
Using the point-slope form again, the equation of the normal line at the point [tex]\((2, 14)\)[/tex] is:
[tex]\[ y - 14 = -\frac{1}{9}(x - 2) \][/tex]
[tex]\[ y - 14 = -\frac{1}{9}x + \frac{2}{9} \][/tex]
[tex]\[ y = -\frac{1}{9}x + \frac{2}{9} + 14 \][/tex]
[tex]\[ y = -\frac{1}{9}x + \frac{2}{9} + \frac{126}{9} \][/tex]
[tex]\[ y = -\frac{1}{9}x + \frac{128}{9} \][/tex]
So, the equation of the normal line is:
[tex]\[ y = -\frac{1}{9}x + \frac{128}{9} \][/tex]
In summary, the equations of the tangent and normal to the curve [tex]\( y = \frac{8}{x} - x + 3x^2 \)[/tex] at the point where [tex]\( x = 2 \)[/tex] are:
- Tangent line: [tex]\( y = 9x - 4 \)[/tex]
- Normal line: [tex]\( y = -\frac{1}{9}x + \frac{128}{9} \)[/tex]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
