Find the gradient of a curve whose parametric equations are x=t^2/2+1 and y=t/4-1 when t=2

Remember that the gradient of a curve is expressed as dy/dx. This can be solved by using the chain rule:

dy/dx = dy/dt*dt/dx. The dt in the denominator of the first term, and the numerator of the second term will cancel. It is also useful to remember that dt/dx is the same as 1/(dx/dt).

Now all we have to do to solve this is find the differential with respect to t of the the two parametric equations. Remember that if we have an equation where there is more than one term (i.e. + or - terms), each term can be differentiated separately and then added together afterwards to give the total differential.

Equation for x

First term: t2/2

Take the power which t is raised by (2) and multiply it by the coefficient of t(1/2), then drop the power by one.

d/dt [t2/2] = t

Second term: +1

The differential of a constant is always equal to 0.

d/dt [1] = 0.

... dx/dt = d/dt [t2/2 + 1] = d/dt [t2/2] + d/dt [1] = t + 0 =t

Equation for y

Repeat the process for y. This is a little tricker since the t is in the denominator of the first term. It is easier to perform the differential if we rewrite the term 4/t as 4t-1. The equation for y is now


Using the same method as before, for the first term:

d/dt [4t-1] = -4t-2

and for the second term:

d/dt [-1] = 0

...dy/dt = d/dt [4t-1-1] = d/dt [4t-1] + d/dt [-1] = -4t-2 + 0 = -4t-2 = -4/t2.

Putting all this together

The equation we need to find the gradient is

dy/dx = dy/dt*dt/dx = dy/dt+1/(dx/dt)

We have already worked out dy/dt and dx/dt. To get dt/dx we just take the reciprocal of dx/dt (that is, switch the denominator and numerator round- in this case the denominator would be 1 as t=t/1).

dt/dx = 1/(dx/dt) = 1/t

Now dy/dx = dy/dt*dt/dx = -4/t2 * 1/t = (-4*1)/(t2*t) = -4/t3.

Now that we have an equation for the gardient, dy/dx, we can simply substiute in our value for t given in the question (t=2).

The gradient at t=2 is therefore:

dy/dx = -4/2= -4/(2*2*2) = -4/8 = -1/2

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