how do i do this.
find the equation for the lines that are tangent and normal to the curve y= (square root of 2)(cosx)at the point (pi/4, 1)
y = √2 cosx
I usually check if the given point is actually on the line
LS = 1
RS = √2 cos(pi/4) = √2(1/√2) = 1 = LS
dy/dx = -√2(sinx)
at (pi/4,1)
dy/dx = -√2(1/√2) = -1
So the slope of the tangent is -1
and the slope of the normal is +1
Take it from there,
every student taking Calculus surely has to know how to find the equation of a line given the slope and a point.
use product rule, then substitute value of x in dy/dx. the obtain then find equation of tangent y=mx+c.
For normal,
gradient of curve*gradient of normal=-1
find gradient of normal from equation and once again obtain normal equation in y=mx+c form
To find the equation for the lines that are tangent and normal to the curve y = √2 * cos(x) at the point (π/4, 1), you can follow these steps:
1. Find the derivative of the function y = √2 * cos(x) with respect to x to determine the slope of the tangent line at any given point on the curve.
2. Determine the slope of the tangent line at the point (π/4, 1) by evaluating the derivative at that point.
3. Use the point-slope form of a line to write the equation for the tangent line using the slope found in step 2 and the coordinates of the given point.
4. The normal line is perpendicular to the tangent line and has a slope that is the negative reciprocal of the tangent line's slope.
5. Use the point-slope form again to write the equation for the normal line using the slope found in step 4 and the coordinates of the given point.
Let's go through these steps one by one:
Step 1: Find the derivative of y = √2 * cos(x):
The derivative of y = √2 * cos(x) can be found using the chain rule. Note that the derivative of cos(x) is -sin(x). Therefore, the derivative of y with respect to x is:
dy/dx = d(√2 * cos(x))/dx
= (√2) * d(cos(x))/dx
= (√2) * (-sin(x))
= -√2 * sin(x)
Step 2: Determine the slope of the tangent line at (π/4, 1):
Evaluate the derivative, -√2 * sin(x), at x = π/4:
-√2 * sin(π/4) = -√2 * (1/√2) = -1
So, the slope of the tangent line at the point (π/4, 1) is -1.
Step 3: Write the equation for the tangent line:
Using the point-slope form of a line, with the slope (-1) and the point (π/4, 1), the equation for the tangent line is:
y - y₁ = m(x - x₁)
y - 1 = -1(x - π/4)
y - 1 = -(x - π/4)
y - 1 = -x + π/4
y = -x + (π/4 + 1)
Thus, the equation for the tangent line is y = -x + (π/4 + 1).
Step 4: Determine the slope of the normal line:
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is -1, the slope of the normal line is 1.
Step 5: Write the equation for the normal line:
Using the point-slope form of a line, with the slope (1) and the point (π/4, 1), the equation for the normal line is:
y - y₁ = m(x - x₁)
y - 1 = 1(x - π/4)
y - 1 = x - π/4
y = x + (π/4 - 1)
Therefore, the equation for the normal line is y = x + (π/4 - 1).
So, the equations for the tangent line and normal line to the curve y = √2 * cos(x) at the point (π/4, 1) are:
Tangent line: y = -x + (π/4 + 1)
Normal line: y = x + (π/4 - 1)
To find the equation for the lines that are tangent and normal to the given curve at the point (pi/4, 1), we need to find the slope of the tangent line and the slope of the normal line.
Step 1: Find the derivative of the function y = √2 * cos(x) with respect to x.
To find the derivative, you can use the chain rule. The derivative of cos(x) is -sin(x), and the derivative of √2 is 0 (since it is a constant).
dy/dx = √2 * (-sin(x)) = -√2 * sin(x)
Step 2: Plug in the x-coordinate (pi/4) into the derivative to find the slope of the tangent line at the point (pi/4, 1).
dy/dx = -√2 * sin(pi/4) = -√2 * (1/√2) = -1
Therefore, the slope of the tangent line is -1.
Step 3: The slope of the normal line is the negative reciprocal of the slope of the tangent line.
The negative reciprocal of -1 is 1.
Step 4: Use the point-slope form of a straight line to find the equations of the tangent and normal lines.
Equation of the tangent line:
Using the point-slope form: y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope of the tangent line.
Plugging in the values, we have:
y - 1 = (-1)(x - pi/4)
Simplifying it, we get:
y - 1 = -x + pi/4
Rearranging it to the slope-intercept form, we get:
y = -x + (pi/4 + 1)
The equation of the tangent line is y = -x + (pi/4 + 1).
Equation of the normal line:
Using the same point-slope form, but with the slope of the normal line (1), we have:
y - 1 = (1)(x - pi/4)
Simplifying it, we get:
y - 1 = x - pi/4
Rearranging it to the slope-intercept form, we get:
y = x + (pi/4 + 1)
The equation of the normal line is y = x + (pi/4 + 1).
That's how you find the equations for the lines that are tangent and normal to the given curve at the point (pi/4, 1).