Find the gradient of the tangent to the curve 3x^2+4y^2 = 7 at the point where x=1 and y>0.?

Question

I don't even know what to do first, get the derivative of the curve?

Answer 1

Funny the way this question is stated. Do you have to find the gradient of 3x^2 + 4y^2 in the point where this formula gives 7 and x = 1, y > 0?

Or are you looking for the slope of the tangent of the curve at that point?

In each case, you need to know y:
3x^2 + 4y^2 = 7 ... use x = 1 ...
3 + 4y^2 = 7
y^2 = 1
y = +1 ... because y > 0 was given ...

The gradient in a point is found by taking derivatives of 3x^2 + 4y^2 with respect to x and to y. You find the vector (6x, 8y), and plugging in x = 1, y = 1 the gradient turns out to be (6, 8) at that point.

The slope of the tangent to the curve can be found by calculating the total derivative of
3x^2 + 4y^2 = 7
6x dx + 8y dy = 0
6x dx = - 8y dy
dy/dx = (6x)/(-8y) = -3/4 x/y
In the point (1,1) the slope is -3/4.

Answer 2

the gradient is:
(6x,8y)
for x=1
3+4y^2=7, so 4y^2=4, so y^2=1, so y=1 or y=-1, but since you are interested in the part where y>0, then y=1
so the gradient at (1,1) is:

(6,8)

Answer 3

well Anon. DutchProf is right.

Good Luck