0
$\begingroup$

The question is: If $y = m\sqrt{1-nx}$, (where m and n are constants) has a tangent line $6x + 2y = 10$ at $x = -1$. Find the values of m and n.

Attempted solution:

I isolated y in both equations and set them equal to get one equation for m in terms of n:

$5 - 3x = m\sqrt{1-nx}$

$m = \frac{8}{\sqrt{1+n}}$

To get the second equation I set the derivatives of the two original equations equal and solved for m:

$-3 = \frac{1}{2}m(1+n)^{\frac{-1}{2}}$

$m = -6\sqrt{1+n}$

But when I try to solve for m, I don't get the right answer. Also, If I put the correct solutions for n and m into my second equation it results in an untrue statement.

How do I find the correct second equation?

The solutions are: $m = 4, n = 3$

$\endgroup$
1
$\begingroup$

As you differentiate it, the gradient should be

$$-3 =- \frac{mn}2 (1+n)^{-\frac12}$$

Your mistake is you forgot to apply chain rule and omitted a $-n$ factor on the RHS.

From the first equation, we have

$$m = \frac{8}{\sqrt{1+n}}$$

Hence we have

$$-3=-\frac{4n}{1+n}$$

and we have

$$3+3n=4n$$

Hence $n=3$.

$\endgroup$
0
$\begingroup$

When you take the derivative of $5 - 3x = m\sqrt{1-nx}$ wrt $x$ you forgot sth in RHS:

That is it should be $-3 = \frac{1}{2}m(1-nx)^{\frac{-1}{2}}(-n)$
then you get $-3 = \frac{1}{2}m(1+n)^{\frac{-1}{2}}(-n)\\ \iff 6=mn(1+n)^{\frac{-1}{2}} \iff 6=\frac {m^2 n} {8} \iff 48=\frac {64n}{1+n} \iff n=3$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.