the equation of a curve is given by formula x=5t^2 & y=3t-1.find the equa.of the tangent to the curve whn t=3.

Answer 1

x=5t^2
y=3t-1

dx/dt = 10t
dy/dt = 3

dy/dx = (dy/dt) / (dx/dt)
=3/(10t)

When t=3, dy/dx = 1/10 = 0.1

When t=3
x=45
y=8

y=mx+c
m=0.1
x=45
y=8

8=0.1*45 + c
c=3.5

y = 0.1x + 3.5
10y = x + 35
x - 10y + 35 = 0

Answer 2

The gradient of the tangent at the point t is:
dy / dx
= ( dy / dt ) / ( dx / dt )
= 3 / 10t

When t = 3, the gradient is 3 / 30 = 1/10.
The line is therefore y = x /10 + c .........(1)
Evaluating x and y from the equation of the curve with t = 3, the tangent passes through the point (5*3^2, 3*3 - 1) = (45,8).
Substituting this in (1):
8 = 45 /10 + c
c = 8 - 9 / 2
c = 7/2.

The equation of the tangent is therefore:
y = x / 10 + 7 / 2
10y = x + 35.

Answer 3

dx/dt = 10 t
dy/dt = 3
dy/dx = 3 / (10 t)
dy/dx = 3 / 30 = 1 / 10
Tangent passes thro` (45, 8)
y - 8 = (1/10).(x - 45)
10y - 80 = x - 45
10y = x + 35
x - 10y + 35 = 0 is acceptable form for equation of tangent when t = 3
May be written as:-
y = (1/10).x + 3.5 if desired.

Answer 4

x = 5t² , y = 3t - 1

dx/dt = 10t, dy/dt = 3

dy/dx = 3/(10t)

when t = 3 , dy/dx = 1/10, x = 45 and y = 8

Equation of tangent is (y - 8) / (x - 45) = 1/10

10y - 80 = x - 45

x - 10y + 35 = 0

Answer 5

x=5t^2
dx/dt=10t

y=3t-1
dy/dt=3

dy/dx=(dy/dt)*(dt/dx)
=3*(1/10t)
=3/10t

gradient of tangent to the curve=3/10(3)
=1/10

x1=5t^2
=5(3)^2
=45

y1=3t-1
=3(3)-1
=8

y-y1=m(x-x1)
y-8=(1/10)(x-45)
y=x/10-45/10+8
y=x/10+35/10
10y=x+35