M1 position vector question

hi, please can anyone help me with part b? I have got up to the 5t^2-30t+50 part but I have no idea what d^2 is. The mark scheme shows completing the square and I don't understand what is 5 and why t is somehow 3... Thanks! Please try to explain it simply.

We talked about this before, remember? It's a maximisation/minimisation problem. You have d^2 as a quadratic and you want to find the minimum point of d^2. There are two ways of finding the minimum point of a quadratic, either by differentiation or by completing the square.

Anyhow, you have $5t^2 - 30t + 50 = 5(t^2 - 6t +10) = 5((t-3)^2 + 1) = 5(t-3)^2 + 5$. Do you still remember how to complete the square?

You have $t^2 - 6t + 10$ so you do $(t - \frac{6}{2})^2$ but that gets you $t^2 - 6t + 9$ so you need to add one.

Hence you have $t^2 - 6t + 10 = (t-3)^2 + 1$.

Hence, you have: $5(t^2 - 6t + 10) = 5(t-3)^2 + 5$

Now, if you remember your C2/C1/Whatever, you should remember that this means your minimum point is at $(3,5)$ - if you don't, then I strongly recommend you go look up completing the square in your textbook.

So the minimum value of $d^2$ occurs when $t=3$ and the minimum value is $d^2 = 5$.

That's a bit like me saying $y = x^2 + 1$ has a minimum at $(0, 1)$ - i.e: the minimum occurs when $x=0$ and the minimum value is $y=1$. Except, in this case, $d^2$ is your $y$ and $x$ is your $t$.

We talked about this before, remember? It's a maximisation/minimisation problem. You have d^2 as a quadratic and you want to find the minimum point of d^2. There are two ways of finding the minimum point of a quadratic, either by differentiation or by completing the square.

Anyhow, you have $5t^2 - 30t + 50 = 5(t^2 - 6t +10) = 5((t-3)^2 + 1) = 5(t-3)^2 + 5$. Do you still remember how to complete the square?

You have $t^2 - 6t + 10$ so you do $(t - \frac{6}{2})^2$ but that gets you $t^2 - 6t + 9$ so you need to add one.

Hence you have $t^2 - 6t + 10 = (t-3)^2 + 1$.

Hence, you have: $5(t^2 - 6t + 10) = 5(t-3)^2 + 5$

Now, if you remember your C2/C1/Whatever, you should remember that this means your minimum point is at $(3,5)$ - if you don't, then I strongly recommend you go look up completing the square in your textbook.

So the minimum value of $d^2$ occurs when $t=3$ and the minimum value is $d^2 = 5$.

That's a bit like me saying $y = x^2 + 1$ has a minimum at $(0, 1)$ - i.e: the minimum occurs when $x=0$ and the minimum value is $y=1$. Except, in this case, $d^2$ is your $y$ and $x$ is your $t$.

Oh yeah, I can never relate the two together, it never occurred to me that it is about minimum points! Anyway cheers again.