how do i find the equaiton of the line tangent to the curve at the point defined by the given value of t.
1. x=t, y= the square root of t, t=1/4
2. x=sec^2t-1, y= tan t, t= -pi/4
I assume you are in calculus.
y= t^.5
y'=.5/t^.5
at t= .25 y'= .5/(.25)^5= .5/.5=1
y= mx + b
y= 1 x + b
put in x,y and solve for b.
To find the equation of the line tangent to a curve at a point defined by a given value of t, you'll need to follow these steps:
1. Determine the derivative of the given curve.
2. Substitute the given value of t into the derivative to find the slope of the tangent line.
3. Use the point-slope form or the slope-intercept form to write the equation of the line.
Now let's go through the steps for each given value of t to find the equations of the tangent lines:
1. For x = t and y = square root of t, t = 1/4:
a. To find the derivative, differentiate each equation with respect to t.
- Differentiating x = t gives dx/dt = 1.
- Differentiating y = sqrt(t) gives dy/dt = (1/2) * (1/sqrt(t)).
b. Substitute the value of t into the derivatives to find the slope of the tangent line.
- Substituting t = 1/4 in dx/dt gives dx/dt = 1.
- Substituting t = 1/4 in dy/dt gives dy/dt = (1/2) * (1/sqrt(1/4)) = 1.
c. The slope of the tangent line is 1.
d. Using the point-slope form, we have y - y₁ = m(x - x₁), where (x₁, y₁) is the given point on the curve.
- Substituting x₁ = t = 1/4 and y₁ = sqrt(t) = sqrt(1/4) = 1/2, we have y - 1/2 = 1(x - 1/4).
e. Simplifying, the equation of the tangent line is y = x - 1/4.
2. For x = sec^2(t) - 1 and y = tan(t), t = -pi/4:
a. To find the derivative, differentiate each equation with respect to t.
- Differentiating x = sec^2(t) - 1 gives dx/dt = 2 * sec(t) * tan(t).
- Differentiating y = tan(t) gives dy/dt = sec^2(t).
b. Substitute the value of t into the derivatives to find the slope of the tangent line.
- Substituting t = -pi/4 in dx/dt gives dx/dt = 2 * sec(-pi/4) * tan(-pi/4) = -2.
- Substituting t = -pi/4 in dy/dt gives dy/dt = sec^2(-pi/4) = 2.
c. The slope of the tangent line is -2.
d. Using the point-slope form, we have y - y₁ = m(x - x₁), where (x₁, y₁) is the given point on the curve.
- Substituting x₁ = sec^2(t) - 1 = sec^2(-pi/4) - 1 = (-1/2) - 1 = -3/2 and y₁ = tan(t) = tan(-pi/4) = -1, we have y - (-1) = -2(x - (-3/2)).
e. Simplifying, the equation of the tangent line is y = -2x + 2.