For the question "Determine the equation of the tangent to the curve y = xtanx at the point with x-coordinate π." how is the answer -πx + y + π2 = 0?
The equation of the tangent line can be represented by the point-slope format which is y-y0=m(x-x0). The point of interest is x=pi.
Since there is no y0 coordinate provided, the tangent line equation will be in the form y=m(x-x0).
To calculate the slope(m), take the derivative of y=xtan(x) by applying the product rule, which would yield y'=m=xsec^2(x)+tan(x). Plug in pi into y' which gives pi so the value of m is pi.
Now plug everything into y=m(x-x0) -> y=pi(x-pi) -> So the equation of the tangent line is y=pix-pi^2
To determine the equation of the tangent to the curve y = xtan(x) at the point with x-coordinate π, you need to find the slope of the tangent line and its y-intercept.
Step 1: Differentiate the equation y = xtan(x) using the product rule.
- To differentiate the term x, you get 1 as its derivative.
- To differentiate the term tan(x), you can use the derivative of tan(x) formula, which is sec^2(x).
- Applying the product rule, the derivative of y = xtan(x) is given by:
dy/dx = 1 * tan(x) + x * sec^2(x)
Step 2: Calculate the slope of the tangent at x = π.
- Substitute x = π into the derivative obtained in step 1 to find the slope of the tangent at x = π.
dy/dx = 1 * tan(π) + π * sec^2(π)
= 1 * 0 + π * (1^2)
= π
So, the slope of the tangent line at x = π is equal to π.
Step 3: Find the y-coordinate of the point on the curve when x = π.
- Substitute x = π into the original equation y = xtan(x).
y = π * tan(π)
Step 4: Use the point-slope form of a linear equation.
- The equation of a line in point-slope form is y - y1 = m(x - x1), where (x1, y1) is a point on the line, and m is its slope. In this case, m = π, x1 = π, and y1 = π * tan(π). Plugging in these values:
y - (π * tan(π)) = π(x - π)
Step 5: Simplify the equation.
- Distribute π on the right side:
y - (π * tan(π)) = πx - π^2
- Add π * tan(π) to both sides to isolate the y-term:
y = πx - π^2 + π * tan(π)
- Use the identity tan(π) = 0:
y = πx - π^2 + 0
- Simplify further:
y = πx - π^2
- Rearrange the terms to match the given answer format:
-πx + y + π^2 = 0
Therefore, the equation of the tangent line to the curve y = xtan(x) at the point with x-coordinate π is -πx + y + π^2 = 0.
To determine the equation of the tangent line to the curve y = xtanx at the point with x-coordinate π, we can use the concept of differentiation.
First, let's find the derivative of the curve y = xtanx. The derivative, dy/dx, represents the slope of the tangent line at any given point on the curve.
To find the derivative, we can use the product rule. Let u = x and v = tanx.
Differentiating both u and v with respect to x:
du/dx = 1 (since the derivative of x with respect to x is 1)
dv/dx = sec^2x (since the derivative of tanx with respect to x is sec^2x)
Now, applying the product rule:
dy/dx = x * dv/dx + v * du/dx
= x * sec^2x + tanx
Next, substitute π for x in dy/dx to find the slope of the tangent line at the point with x-coordinate π:
m = dy/dx at x = π = π * sec^2(π) + tan(π)
Simplifying this expression:
sec(π) = 1 (since sec(π) is the reciprocal of cos(π), and cos(π) = -1)
tan(π) = 0 (since tan(π) is the sine-cosine ratio, and sin(π) = 0)
m = π * (1)^2 + 0
m = π + 0
m = π
Now that we have the slope of the tangent line, we need to find the y-intercept. We can use the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is the point of tangency.
Since the point of tangency has x-coordinate π, we substitute π for x and solve for y:
y - 0 = π(x - π)
y = πx - π^2
Finally, rearrange the equation to the standard form:
πx - y - π^2 = 0
Therefore, the equation of the tangent to the curve y = xtanx at the point with x-coordinate π is -πx + y + π^2 = 0.