Find an equation of the tangent line to the curve at the given point. y = sin(sin x), (3π, 0)?
y(x) = sin(sin x), so y(3π) = sin(sin 3π) = sin(0) = 0, so (3π, 0) is on the curve y'(x) = cos(x) cos(sin(x)), so y'(3π) = cos(3π) cos(sin(3π)) = cos(3π) cos(0) = -1 × 1 = -1 y = -(x-3π) - 0, which is x + y = 3π ⇦ ⇦ ⇦ ??????
Find an equation of the tangent line to the curve at the given point? Calculus help?
y' = sec x tan x = sec π/6 tan π/6 = (1/2)/(3/4) = 2/3 Answer: y - 2sqrt(3)/3 = (2/3)(x - π/6)
Find an equation of the tangent line to the curve at the given point. y = 9e^x cos x, (0, 9)?
d/dx(e^x)=e^x and d/dx(cosx)=-sinx I use the Product rule in the derivation. dy/dx=9e^xcosx-9e^xsinx At (0,9) dy/dx=9e^0=9 Then the tangent through the point has slope 9. y=9x+c is the Equation, and c is a constant number. As (0,9) is on the line, we have this relation: 9=9*0+c c=9 Equation for the tangent: y=9x+9
Find the equation of the tangent line to the curve at the given point?
Okay, since you already know the derivative, you found the equation of the slope. So you take the derivative and plug in the given x value to find the value of the slope. y'=3x^2-3 y'=3(3)^2-3 which equals 24 Now since you know what the value of the slope is (m=slope) and you have a x and y value (3,19), then you can plug it into the slope intercept form to find the value of b. The equation is y=mx+b. 19=(24)(3)+b 19=72+b b=-53 Now since you have the value of b and m, you just put it into y=mx+b. y=24x-53 is the equation of the tangent line.
Find an equation of the tangent line to the curve at the given point.?
The slope of a tangent line to a function f(x) at point x is given by the derivative of the function, f'(x). f(x) = x^3 - 2x + 1 f'(x) = 3x^2 - 2 The slope of the tangent line of f(x) = x^3 - 2x + 1 at x = 2 is f'(2) = [ 3 * (2)^2 ] - 2 = 12 - 2 = 10 Then you can use the point-slope form of the equation for a line to find the equation of the tangent line at (2,5) . (y - y1) = m(x - x1) (y - 5) = 10(x - 2) y - 5 = 10x - 20 y = 10x - 15 . . . . . <--- that's the equation of the desired tangent line ------------ I graphed the function and the tangent line equation to verity that it is correct. Here's an online graphing calculator : http://www.meta-calculator.com/online/
Find an equation of the tangent line to the curve at the given point. y = sin(sin x), (4π, 0)?
y = sin(sinx) y' = cos(sinx) * cosx y'(4π) = cos(sin(4π)) * cos(4π) = cos(0) * 1 = 1 Tangent line has slope 1 and passes through point (4π,0) y − 0 = 1 (x − 4π) y = x − 4π
Find an equation of the tangent line to the curve at t=π/2, x=t sin t and y=t cos t. Can you solve this tangent problem?
The slope of tangent =dy/dxx=tsintdx/dt =tcost +sinty=tcostdy/dt =-tsint +costdy/dx=cost-tsint/tcost+sintdy/dx at t=π/2dy/dx=-π/2Value of x at t=π/2 ,x=π/2Value of y at t=π/2 ,y=0Equation of tangent line:-y-y₁=m(x-x₁)y=-π/2(x-π/2)
How can we find the equation of a unit tangent at a point on a curve?
I’m not sure what you mean by unit tangent, but here is how to find the equation of the tangent at a point on a curve.Let the curve be y = f(x) and the point (a,b)Find the derivative f´(x)Now find f´(a)The equation of the tangent is y - b = f´(a) (x - a)
What is the equation of the tangent line to the curve?
[math]f’(x_0)(x-x_0)+f(x_0)[/math].Where [math]f(x)[/math] is the function of the curve,[math]x_0[/math] is the x-coordinate of the point at which the curve and the tangent line touch each other, and[math]f’(x)[/math] is the derivative of the function.