Question

Find the equation of the straight line which passes through the point (3,1) and has a gradient of -4. Given that this line is parallel to the tangent to the curve y=k/x^2 at the point where x=2 calculate the value of k

Answer 1

The gradient cannot be a single number (-4), since it is a vector quantity.

Answer 2

gradient = slope = m= -4
x1,y1 = 3,1

so using the slope point form of the equation of line,
\(\large y-y_1= m(x-x_1)\)

y- 1 = (-4) (x-3)

you can simplify this to get your equation of line :)

Answer 3

"tangent to the curve y=k/x^2 at the point where x=2"
has the slope same as derivative of y at x=2

so calculate dy/dx first,
y= k/x^2 = kx^-2

dy/dx = k (-2) x^(-1)

now the slope m = dy/dx at x= 2 = k (-2) * 2^(-1) = -k

since this tangent is parallel to your line, their slopes are same
so, -k = -4
gives you \(\large \color{red}{k=4}\)
:)

Answer 4

When i looked back at my book it appears to be that k=16 and y+4k=13

Answer 5

oh, i did my derivative incorrectly.
y = kx^-2
so,
dy/dx = k (-2) x^-3
- 4 = -2k (2)^-3
4 = 2k/8
k = 32/2
k=16

Answer 6

Thank you so much for your help :D

Answer 7

you're welcome ^_^