# Math Help - Calculus problem

1. ## Calculus problem

Here is the exact problem that my teacher gave me. I don't necessarily need the answer i just need to know how i would go about solving this problem. i have tried for two days but i have made no progress. click on the picture to view the problem.

2. This might be messy.
The fundamental equation:The equation of a line passing through $(x_0,y_0)$ with slope $m$ is $y-y_0=m(x-x_0)$.

Thus, find a point on this parabola ( $x_0$) such as it is a tangent and passes through $(116,0)$ because this is the point of the tower. I am going to use the fundamental equation for this problem which I stated in the previous paragraph. But to use it you need to know the slope at point $x_0$ and the y-coordinate of that point $y_0$. But it slope is its derivative there thus it is. $m=\frac{x_0}{-200}$ and $y_0=100-\frac{{x_0}^2}{400}$
Thus, the equation of the tangent for any point $x_0$ is $y-y_0=m(x-x_0)$ substitute your values to get watch the negative signs
$y-100+\frac{{x_0}^2}{400}=\frac{x_0}{-200}(x-x_0)$
Simplify,
$y-100+\frac{{x_0}^2}{400}=-\frac{xx_0}{200}+\frac{{x_0}^2}{200}$
Simplify further,
$y=-\frac{xx_0}{200}+\frac{{x_0}^2}{400}+100$

Now the line passes through $(0,116)$ thus when $x=0$ then $y=116$ substitute that infromation into the equation
$116=\frac{{x_0}^2}{400}+100$
Thus,
$6400=(x_0)^2$
$x_0=\pm 80$
(which is true you can draw this tangent through two points one on the right and one on the left). But the picture is positioned on the right thus, $x_0=-80$
Substitute that into your tangent equation thus,
$y=-\frac{2}{5}x+116$
Q.E.D.

3. The second part is much more easier. Just subtract the distance of Q from the origin from P from the origin. That means find the x-intercepts. That is when $y=0$ Thus,
$0=100-\frac{x^2}{400}$
$0=-\frac{2}{5}x+116$
For the first equation we get, $x=200$.
For the second equation we get, $x=290$
Thus, the answer is $290-200= 90ft$

4. your answers were very similar but not exactly what i got when i used my calculator. the equation that my calculator produced when i checked it was y=-.375x+116 (-.375 is very close to negative sqaure root of two over five which is what you got) i double checked and it is correct on my calculator. i am not saying you are wrong im just saying that you and my calculator disagree

5. Originally Posted by arollyson
Here is the exact problem that my teacher gave me. I don't necessarily need the answer i just need to know how i would go about solving this problem. i have tried for two days but i have made no progress. click on the picture to view the problem.
We are interested in finding the tangent to the parabola which passes
through the point $(0,116)$. We don't need calculus to do
this, just to observe that if $y=mx+c$ is the equation of the
line, then it intersects the parabola in but a single point.

The tangent passes through $(0,116)$, so we can give a value
$c=116$, and so the equation of the tangent is $y=mx+116$.

The $x$ coordinates of the points at which this line meets the
parabola are the roots of:

$mx+116=100-\frac{x^2}{400}$

or rearranging:

$\frac{x^2}{400}+mx+16=0$.

This quadratic has a single root if the discriminant is zero (thats the bit
under the square root in the quadratic formula - this is the condition
for the quadratic to be the square of a single linear factor),
so for a tangent:

$m^2-4 \frac{1}{400}16=0$,

or:

$m^2=\frac{4}{25}$.

So $m=\pm \frac{2}{5}$, but from your diagram it is clear that
we want the tangent with negative slope so we want $m=- \frac{2}{5}$,
and so the equation of the tangent is:

$y=-\frac{2}{5}x+116$.

RonL

6. Originally Posted by arollyson
your answers were very similar but not exactly what i got when i used my calculator. the equation that my calculator produced when i checked it was y=-.375x+116 (-.375 is very close to negative sqaure root of two over five which is what you got) i double checked and it is correct on my calculator. i am not saying you are wrong im just saying that you and my calculator disagree
I'm curious how did you do this on your calculator?

RonL

7. Originally Posted by arollyson
Here is the exact problem that my teacher gave me. I don't necessarily need the answer i just need to know how i would go about solving this problem. i have tried for two days but i have made no progress. click on the picture to view the problem.
Here is one way.

a) The equation of the tangent line TQ.

If you know a point in a line, and you also know the slope of the same line, then you can get the equation of the line in its point-slope form:
(y -y1) = m(x -x1)
Where (x1,y1) is the known point, and m is the slope.

So we find a known point in TQ, and the slope of TQ.

Point T is easy to find if we know height of the top of the hill, or R, from the x-axis. That R is when x = 0 in the parabola,
y = 100 -(x^2)/400
y = 100 -(0^2)/400 = 100
So, R is 100 ft above the x-axis. Hence, point T is (100+16) = 116 ft above the x-axis.
Since at T, x=0 also, then point T is (0,116). That is a known point in TQ.

Since TQ is tangent to the parabola, then the slope of TQ is the value of the first derivative of the parabola at the point of tangency.
y = 100 -(x^2)/400
Differentiate both sides with respect to x,
dy/dx = -x/200 ----------the slope of the tangent line to the parabola at any point.
That is also the slope of TQ because TQ is tangent to the parabola.
So, slope m = -x/200.

The point-slope form of the equation of TQ then is:
(y -116) = (-x/200)(x -0)
Simplifying that,
y -116 = (-x^2)/200
y = (-x^2)/200 +116 --------(i)
That is not yet the equation of the tangent line, because of the x^2.
A line, straight line, tangent line, is linear only, or the highest degree of the variable is one only.

At the point of tangecy, the y of the parabola is the same as the y of TQ, so,
100 -(x^2)/400 = (-x^2)/200 +116
-(x^2)/400 +(x^2)/200 = 116 -100
(x^2)/400 = 16
x^2 = 16*400
Take the square roots of both sides,
x = +,-4*20 = +,-80
Since the point of tangency is to the right of the y-axis, then x = 80.
Substitute that into the m = -x/200,
m = -80/200
m = -2/5

Therefore, the equation of TQ is:
y -116 = (-2/5)(x -0)
y = -(2/5)x +116 --------------answer, in its y=mx+b form.

--------------------------------
b) The distance between points P and Q.

Point P is when y=0 in the parabola.
y = 100 -(x^2)/400
0 = 100 -(x^2)/400
(x^2)/400 = 100
x^2 = 100*400
x = +,-10*20 = +,-200
In the figure, x is to the right of the y-axis, so x = 200.
And point P is (200,0).

Point Q is when y=0 in line TQ.
y = -(2/5)x +116
0 = -(2/5)x +116
(2/5)x = 116
x = 116/(2/5) = 116*(5/2) = 58*5 = 290
So point Q is (290,0)

Therefore, the distance between P and Q is 290 -200 = 90 ft. ------answer.

8. Sorry arollyson I made a mistake yesterday. It is now fixed.
I edited my post correctly this time you may read it.

9. i didnt really find the correct answer with my calculator it was more of a guess and check until it looked right on the graph. but thanks to everyone who helped, i dont know why i was having so much trouble figuring that out. thanks again