Question

Find an equation for the line tangent to the curve at the point defined by the given value of t. Also, find the value of d^2y/dx^2 at this point.
x=sin^2t-1, y=csc(t); t=-pi/6

Answer 1

Answer:

Differentiation will give you the gradient for the tangent at any point, and you use the product rule whenever a function can be thought of as two functions multiplied together.

If

f

(

x

)

=

g

(

x

)

×

h

(

x

)

then

f

'

(

x

)

=

g

'

(

x

)

h

(

x

)

+

g

(

x

)

h

'

(

x

)

so if

y

=

x

×

sin

x

then

d

y

d

x

=

1

×

sin

x

+

x

×

cos

x

=

sin

x

+

x

cos

x

We know that

x

=

π

2

, so the gradient is

m

=

sin

(

π

2

)

+

π

2

cos

(

π

2

)

=

1

+

π

2

×

0

=

1

Therefore, we can say that

y

=

m

x

+

c

y

=

(

1

)

x

+

c

y

=

x

+

c

So all we really need to find now is the value for

c

, the

y

intercept. We do this by working out a point

(

x

,

y

)

on the graph. We are already given that

x

=

π

2

, so

y

=

x

sin

x

=

π

2

sin

(

π

2

)

=

π

2

×

1

=

π

2

∴

(

x

,

y

)

=

(

π

2

,

π

2

)

Now we substitute this into the equation we already have for the tangent,

y

=

x

+

c

,

(

x

,

y

)

=

(

π

2

,

π

2

)

π

2

=

π

2

+

c

c

=

π

2

−

π

2

=

0

∴

y

=

x

+

c

=

x

+

(

0

)

=

x

which means the tangent to the curve

y

=

x

sin

x

at

(

π

2

,

π

2

)

is simply

y

=

x

.