help me with calculus?

Question

Write the equation of the line tangent to the curve described by x=t and y=square root of t at the point where t=1/4.

show work.

Answer 1

dy/dx = (dy/dt)/(dx/dt)
dy/dt = 1/(2√t)
dx/dt = 1

dy/dx = 1/(2√t)
x=t
y=√t

Substituting t=1/4:

dy/dx = 1
x = 1/4
y = 1/2

Using point-slope form:

y-1/2 = 1(x-1/4)

Simplifying:

y=x+1/4

Answer 2

x = t
y = √t

When t = ¼, x = ¼, y = ½
dy/dx = dy/dt * dt/dx
= 1/(2√t) * 1
= 1/(2√t)
= 1 when t = ¼
= slope of required tangent

Tangent is of form
y = mx + b
m = 1
So y = x + b
When x = ¼, y = ½
so ½ = ¼ + b
b = ¼

So y = x + ¼

Answer 3

x=t
y=sqrt(t)=sqrt(x)
dy/dx=1/(2*sqrt(x))
at t=0.25, x=0.25
dy/dx=1(2*sqrt(0.25))=1