Again please help with differentiation?

Question

The Q states: The curve C has equation y = 3x^3/2 - 32/x

(A) Find the equation of the tangent to C at the point where x = 4

(B) Find the equation of the normal to C at the point where x = 4

Answer 1

(a) y = 3x^[3/2] - 32/x

To find the equation of the tangent, we must first solve for the derivative. Let's simplify that a bit to separate constants from variables. The reason why we do this is because we can ignore constants when taking the derivative.

y = 3x^(3/2) - 32(1/x)

Taking the derivative, we have

dy/dx = 3(3/2)x^(1/2) - 32(-1/x^2)

Cleaning this up a bit, we get

dy/dx = (9/2)x^(1/2) + 32/x^2

We want to find the equation of the tangent line at x = 4. Thus, we plug in x = 4, and that will be our slope.

m = (9/2)(4)^(1/2) + 32/(4)^2
m = (9/2)(2) + 32/16
m = 9 + 2
m = 11

Note that when x = 4,
y = 3(4)^(3/2) - 32/4
y = 3(8) - 8 = 16

So we want to find the slope at (4,16).

We find the equation of the line using our slope formula, with
(4, 16) and (x, y) being our (x1, y1) and (x2, y2).

(y2 - y1) / (x2 - x1) = m
(y - 16) / (x - 4) = 11
(y - 16) = 11(x - 4)
(y - 16) = 11x - 44
y = 11x - 28

(b) To solve for the normal, this is essentially the line perpendicular to y = 11x - 28. Remember that perpendicular means "negative reciprocal" slope.

Since our slope for the above line is 11, our negative reciprocal of 11 would be -1/11. Plugging this into our slope formula

(y - 16)/(x - 4) = -1/11
11(y - 16) = (-1)(x - 4)
11(y - 16) = -x + 4
y - 16 = (-1/11)x + 4/11
y = (-1/11)x + 4/11 + 16
y = (-1/11)x + 4/11 + 176/11
y = (-1/11)x + 180/11

Answer 2

Part a)
y = f(x) = 3x^3/2 - 32x^(-1)
f(4) = 3x4^(3/2) - 32/4^(-1)
f(4) = 3x8-8 =16
Tangent passes thro` (4,16)

dy/dx = f `(x) = (9/2)x^(1/2) + 32x^(-2)
f ` (4) = 9/2 x 4^(1/2) + 32/4²
f `(4) = 9 + 2 = 11= m (the gradient at (4,16)

Equation of tangent with m=11 that passes thro` (4,16) is given by:-
y-16=11(x-4)
y=11x - 44 +16
y= 11x -28

Part b)
normal to tangent at (4,16) has gradient -1/11
Equation of normal is:
y-16= -1/11(x-4)
y= (-1/11)x + (4/11) +16
y=(-1/11)x + 180/11

Answer 3

y = 3x^3/2 - 32/x
dy/dx=(9/2)x^(1/2)+32/x^2
(dy/dx)(x=4)=9+2=11
(dy/dx)1=11
at x1=4,
y1=24-8=16

(A)the equation of the tangent to C
at the point where x = 4 is

y-y1=(dy/dx)1(x-x1)
y-16=11(x-4)=11x-44

>>>y-11x+28=0

(B)the equation of the normal to C
at the point where x = 4 is

y-y1={-1/(dy/dx)1}(x-x1)
y-16=-1/11(x-4)
11y-176=-x+4

>>>11y+x-180=0

i hope that this helps

Answer 4

1)find first derivative

2)sub 4 to find slope

3)-1/m=normal
-----------------------------------------------------

1) y'=9/2x^1/2+32/x^2----x=4

=9/2(2)+2=11

2)y=11(x-4)+16

3) -1/11

y= -1/11(x-4)+16