Assuming that the equation defines x and y implicitly as differentiable functions xequalsf(t), yequalsg(t), find the slope of the curve xequalsf(t), yequalsg(t) at the given value of t. x cubed plus 4 t squaredequals37, 2 y cubed minus 2 t squaredequals110, tequals3

Hence, the slope ,

Step-by-step explanation:

We need to find the slope, i.e. .

and all the functions are in terms of .

So this looks like a job for the 'chain rule', we can write:

Given the functions

and

we can differentiate them both w.r.t to

first we'll derivate Eq(B) to find dx/dt

we can also rearrange Eq(B) to find x in terms of t , . This is done so that is only in terms of t.

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

now let's differentiate Eq(C) to find dy/dt

rearrange Eq(C), to find y in terms of t, that is . This is done so that we can replace y in to make only in terms of t

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

Finally we can plug all of our values in Eq(A)

but remember when plugging in the values that is being multiplied with and NOT , so we have to use the reciprocal!

our slope is equal to

391.6 is the correct answer